GIFT  OF 


INTRODUCTION  TO 


ANALYTIC    GEOMETRY 


BY 
PEECEY  R  SMITH,  PH.D. 

»N 

PROFESSOR  OF  MATHEMATICS  IN  THE  SHEFFIELD  SCIENTIFIC  SCHOOL 
YALE  UNIVERSITY 


AND 


AKTHUB  SULLIVAN  GALE,  PH.D. 

ASSISTANT  PROFESSOR  OF  MATHEMATICS  IN 
THE  UNIVERSITY  OF  ROCHESTER 


GINN  &  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


COPYRIGHT,  1904, 1905,  BY 
ARTHUR  SULLIVAN  GALE 

ALL  BIGHTS  RESERVED 

65.8 


GINN   &   COMPANY  •  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PEE FACE 


In  preparing  this  volume  the  authors  have  endeavored  to  write 
a  drill  book  for  beginners  which  presents,  in  a  manner  conform- 
ing with  modern  ideas,  the  fundamental  concepts  of  the  subject. 
The  subject-matter  is  slightly  more  than  the  minimum  required 
for  the  calculus,  but  only  as  much  more  as  is  necessary  to  permit 
of  some  choice  on  the  part  of  the  teacher.  It  is  believed  that  the 
text  is  complete  for  students  finishing  their  study  of  mathematics 
with  a  course  in  Analytic  Geometry. 

The  authors  have  intentionally  avoided  giving  the  book  the 
form  of  a  treatise  on  conic  sections.  Conic  sections  naturally 
appear,  but  chiefly  as  illustrative  of  general  analytic  methods. 

Attention  is  called  to  the  method  of  treatment.  The  subject  is 
developed  after  the  Euclidean  method  of  definition  and  theorem, 
without,  however,  adhering  to  formal  presentation.  The  advan- 
tage is  obvious,  for  the  student  is  made  sure  of  the  exact  nature 
of  each  acquisition.  Again,  each  method  is  summarized  in  a  rule 
stated  in  consecutive  steps.  This  is  a  gain  in  clearness.  Many 
illustrative  examples  are  worked  out  in  the  text. 

Emphasis  has  everywhere  been  put  upon  the  analytic  side, 
that  is,  the  student  is  taught  to  start  from  the  equation.  He  is 
shown  how  to  work  with  the  figure  as  a  guide,  but  is  warned  not 
to  use  it  in  any  other  way.  Chapter  III  may  be  referred  to  in 
this  connection. 

The  object  of  the  two  short  chapters  on  Solid  Analytic  Geom- 
etry is  merely  to  acquaint  the  student  with  coordinates  in  space 

iii 

20206fi 


iv  PREFACE 

and  with  the  relations  between  surfaces,  curves,  and  equations  in 
three  variables. 

Acknowledgments  are  due  to  Dr.  W.  A.  Granville  for  many 
helpful  suggestions,  and  to  Professor  E.  H.  Lockwood  for  sugges- 
tions regarding  some  of  the  drawings. 


NEW  HAVEN,  CONNECTICUT 
January,  1905 


CONTENTS 


CHAPTER  I 

REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY 
SECTION  PAGE 

1.  Numbers 1 

2.  Constants      ...........  1 

3.  The  quadratic.    Typical  form            .......  2 

4.  Special  quadratics 4 

5.  Cases  when  the  roots  of  a  quadratic  are  not  independent          .        .  5 

6.  Variables 9 

7.  Equations  in  several  variables            .         .         .         .         .         .         .  9 

8.  Functions  of  an  angle  in  a  right  triangle       .....  11 

9.  Angles  in  general 11 

10.  Formulas  and  theorems  from  Trigonometry          ....  12 

11.  Natural  values  of  trigonometric  functions         .         .         .         .         .  14 

12.  Rules  for  signs       ..." 15 

13.  Greek  alphabet 15 


CHAPTER  II 
CARTESIAN  COORDINATES 

14.  Directed  line 16 

15.  Cartesian  coordinates 17 

16.  Rectangular  coordinates 18 

17.  Angles 21 

18.  Orthogonal  projection 22 

19.  Lengths 24 

20.  Inclination  and  slope 27 

21.  Point  of  division 31 

22.  Areas    .  35 

23.  Second  theorem  of  projection 40 


vi  CONTENTS 


CHAPTER  III 

THE  CURVE  AND  THE  EQUATION 

SECTION  PAGE 

24.  Locus  of  a  point  satisfying  a  given  condition 44 

25.  Equation  of  the  locus  of  a  point  satisfying  a  given  condition      .  44 

26.  First  fundamental  problem 46 

27.  General  equations  of  the  straight  line  and  circle          .         .         .  50 

28.  Locus  of  an  equation  ^ 52 

29.  Second  fundamental  problem 53 

30.  Principle  of  comparison  .........  55 

31.  Third  fundamental  problem.    Discussion  of  an  equation    .        .  60 

32.  Symmetry .         .         .         .         .65 

33.  Further  discussion 66 

34.  Directions  for  discussing  an  equation      .         .         .         .         .         .67 

35.  Points  of  intersection  t 69 

36.  Transcendental  curves  72 


CHAPTER  IV 
THE  STRAIGHT  LINE  AND  THE  GENERAL  EQUATION  OF  THE  FIRST  DEGREE 

37.  Introduction    .         .         . 76 

38.  The  degree  of  the  equation  of  a  straight  line       ....  76 

39.  The  general  equation  of  the  first  degree,  Ax  -f  By  +  C  =  0    .         .  77 

40.  Geometric  interpretation  of  the  solution  of  two  equations  of  the 

first  degree 80 

41.  Straight  lines  determined  by  two  conditions 83 

42.  The  equation  of  the  straight  line  in  terms  of  its  slope  and  the  coordi- 

nates of  any  point  on  the  line 86 

43.  The  equation  91  the  straight  line  in  terms  of  its  intercepts      .         .  87 

44.  The  equation/of  the  straight  line  passing  through  two  given  points  88 

45.  The  normal  form  of  the  equation  of  the  straight  line      ...  92 

46.  The  distance  from  a  line  to  a  point     ....  96 

47.  The  angle  which  a  line  makes  with  a  second  line    ....  100 

48.  Systems  of  straight  lines      .         . 104 

49.  The  system  of  lines  parallel  to  a  given  line 107 

50.  The  system  of  lines  perpendicular  to  a  given  line        .         .         .  108 
61.  The  system  of  lines  passing  through  the  intersection  of  two  given 

lines                                                                  HO 


CONTENTS  vii 

CHAPTER  V 
THE  CIRCLE  AND  THE  EQUATION  a;2  +  y2  +  Dx  +  Ey  +  F  =  0 

SECTION  PAGE 

52.  The  general  equation  of  the  circle  ...  .115 

53.  Circles  determined  by  three  conditions 116 

54.  Systems  of  circles 120 

CHAPTER  VI 

POLAR  COORDINATES 

55.  Polar  coordinates 125 

56.  Locus  of  an  equation 126 

67.  Transformation  from  rectangular  to  polar  coordinates    .         .         .  130 

58.  Applications 132 

59.  Equation  of  a  locus 133 

CHAPTER  VII 

TRANSFORMATION  OF  COORDINATES 

60.  Introduction        .         ...        . 136 

61.  Translation  of  the  axes 136 

62.  Rotation  of  the  axes .        .        .  138 

63.  General  transformation  of  coordinates 139 

64.  Classification  of  loci- 140 

65.  Simplification  of  equations  by  transformation  of  coordinates           .  141 

66.  Application  to  equations  of  the  first  and  second  degrees      .         .  144 

CHAPTER  VIII 

CONIC  SECTIONS  AND  EQUATIONS  OF  THE  SECOND  DEGREE 

67.  Equation  in  polar  coordinates          .         .         .         .         .         .         .  149 

68.  Transformation  to  rectangular  coordinates          ....  154 

69.  Simplification  and  discussion  of  the  equation  in  rectangular  coordi- 

nates.   The  parabola,  e  =  1          .         .         .         .         :         .         .154 

70.  Simplification  and  discussion  of  the  equation  in  rectangular  coordi- 

nates.   Central  conies,  e  ^  1 158 

71.  Conjugate  hyperbolas  and  asymptotes 165. 

72.  The  equilateral  hyperbola  referred  to  its  asymptotes  .         .         .  167 


viii  CONTENTS 

SECTION  PAGE 

73.  Focal  property  of  central  conies 168 

74.  Mechanical  construction  of  conies 168 

75.  Types  of  loci  of  equations  of  the  second  degree       ....  170 

76.  Construction  of  the  locus  of  an  equation  of  the  second  degree     .  173 

77.  Systems  of  conies 176 

CHAPTER  IX 

TANGENTS  AND  NORMALS 

78.  The  slope  of  the  tangent      ........  180 

79.  Equations  of  tangents  and  normals 182 

80.  Equations  of  tangents  and  normals  to  the  conic  sections     .         .  184 

81.  Tangents  to  a  curve  from  a  point  not  on  the  curve          .        .        .  186 

82.  Properties  of  tangents  and  normals  to  conies       ....  188 
y  83.  Tangent  in  terms  of  its  slope 192 

CHAPTER  X 
CARTESIAN  COORDINATES  IN  SPACE 

84.  Cartesian  coordinates 196 

85.  Orthogonal  projections.     Lengths 198 

CHAPTER  XI 

SURFACES,  CURVES,  AND  EQUATIONS 

86.  Loci  in  space 201 

87.  Equation  of  a  surface.    First  fundamental  problem         .         .         .  201 

88.  Equations  of  a  curve.    First  fundamental  problem      .         .         .  202 

89.  Locus  of  one  equation.    Second  fundamental  problem            .        .  205 

90.  Locus  of  two  equations.    Second  fundamental  problem      .         .  205 

91.  Discussion    of   the    equations   of    a   curve.     Third    fundamental 

problem 206 

92.  Discussion   of  the   equation   of   a   surface.     Third   fundamental 

__ problem 207 

93.  Plane  and  straight  line             210 

94.  The  sphere 211 

95.  Cylinders         .         .         .         .      » 213 

96.  Cones 214 

97.  Non-degenerate  quadrics          .         .         .        .         .         .         .         .215 


UNIVERSITY', 

OF 


ANALYTIC  GEOMETRY 

CHAPTER   I 
REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY 

1.  Numbers.    The  numbers  arising  in  carrying  out  the  opera- 
tions of  Algebra  are  of  two  kinds,  real  and  imaginary. 

A  real  number  is  a  number  whose  square  is  a  positive  number. 
Zero  also  is  a  real  number. 

A  pure  imaginary  number  is  a  number  whose  square  is  a  nega- 
tive number.  Every  such  number  reduces  to  the  square  root  of 
a  negative  number,  and  hence  has  the  form  b  V—  1,  where  b  is  a 
real  number,  and  (V—  I)2  =—1. 

An  imaginary  or  complex  number  is  a  number  which  may  be 
written  in  the  form  a  +  b  V—  1,  where  a  and  b  are  real  numbers, 
and  b  is  not  zero.  Evidently  the  square  of  an  imaginary  number 
is  in  general  also  an  imaginary  number,  since 

(a  +  b  V^T)2  =  a2  -  b2  +  2  ab  V^l, 
which  is  imaginary  if  a  is  not  equal  to  zero. 

2.  Constants.    A  quantity  whose  value  remains  unchanged  is 
called  a  constant. 

Numerical  or  absolute  constants  retain  the  same  values  in  all 
problems,  as  2,  —  3,  Vl,  TT,  etc. 

Arbitrary  constants,  or  parameters,  are  constants  to  which  any 
one  of  an  unlimited  set  of  numerical  values  may  be  assigned,  and 
these  assigned  values  are  retained  throughout  the  investigation. 

Arbitrary  constants  are  denoted  by  letters,  usually  by  letters  from  the 
first  part  of  the  alphabet.  In  order  to  increase  the  number  of  symbols  at  our 

1 


2  ANALYTIC   GEOMETRY 

disposal,  it  is  convenient  to  use  primes  (accents)  or  subscripts  or  both.  For 
example : 

Using  primes, 

a' (read  "a  prime  or  a  first"),  of'  (read  "a  double  prime  or  a  second"), 
a'"  (read  "a  third"),  are  all  different  constants. 

Using  subscripts, 

&! (read  "  6  one"),  6a  (read  "  6  two"),  are  different  constants. 

Using  both, 

c/(read  "c  one  prime"),  c8"  (read  "c  three  double  prime"),  are  different 
constants. 

3.  The  quadratic.  Typical  form.  Any  quadratic  equation 
may  by  transposing  and  collecting  the  terms  be  written  in  the 
Typical  Form 

(1)  Ax2  +  Bx  +  C  =  0, 

in  which  the  unknown  is  denoted  by  x.  The  coefficients  A,  B,  C 
are  arbitrary  constants,  and  may  have  any  values  whatever, 
except  that  A  cannot  equal  zero,  since  in  that  case  the  equation 
would  be  no  longer  of  the  second  degree.  C  is  called  the  con- 
stant term. 

The  left-hand  member 

(2)  Ax*  +  Bx  +  C 

is  called  a  quadratic,  and  any  quadratic  may  be  written  in  this 
Typical  Form,  in  which  the  letter  x  represents  the  unknown. 
The  quantity  B2  —  4  A  C  is  called  the  discriminant  of  either  (1) 
or  (2),  and  is  denoted  by  A. 

That  is,  the  discriminant  A  of  a  quadratic  or  quadratic  equa- 
tion in  the  Typical  Form  is  equal  to  the  square  of  the  coefficient 
of  the  first  power  of  the  unknown  diminished  by  four  times  the 
product  of  the  coefficient  of  the  second  power  of  the  unknown 
by  the  constant  term. 

The  roots  of  a  quadratic  are  those  numbers  which  make  the 
quadratic  equal  to  zero  when  substituted  for  the  unknown. 

The  roots  of  the  quadratic  (2)  are  also  said  to  be  roots  of  the 
quadratic  equation  (1).  A  root  of  a  quadratic  equation  is  said 
to  satisfy  that  equation. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY 

In  Algebra  it  is  shown  that  (2)  or  (1)  has  two  roots,  Xi  and 
obtained  by  solving  (1),  namely, 


(3) 

Adding  these  values,  we  have 

B        ^Vw*  \& 

(4)  x  -f-  x  =  —  —  • 

Multiplying  gives 

(5)  XiXz  =  — 

'A 

Hence 

Theorem  I.  T^e  sum  of  the  roots  of  a  quadratic  is  equal  to  the 
coefficient  of  the  first  power  of  the  unknown  with  its  sign  changed 
divided  by  the  coefficient  of  th&  second  power. 

The  product  of  the  roots  equals  the  constant  term  divided  by  the 
coefficient  of  the  second  power. 

The  quadratic  (2)  may  be  written  in  the  form 

(6)  Ax*  +  Bx  +  C  =*A(x  -  x,}  (x  -  *2), 

as   may  be  readily   shown  by   multiplying  out  the  right-hand 
member  and  substituting  from  (4)  and  (5). 

For  example,  since  the  roots  of  3  a;2  —  4  a;  +  1  =  0  are  1  and  £,  w$  have  iden- 
tically 3  32-43  +  1  =  3*(x  -  1)  (x  -  *). 

The  character  of  the  roots  xl  and  xz  as  numbers  (§  1)  when  the 
coefficients  A,  B,  C  are  real  numbers  evidently  depends  entirely 
upon  the  discriminant.  This  dependence  is  stated  in 

Theorem  II.  If  the  coefficients  of  a  quadratic  are  real  numbers, 
and  if  the  discriminant  be  denoted  by  A,  then 

when  A  is  positive  the  roots  are  real  and  unequal; 
•4/i  when  A  is  zero  the  roots  are  real  and  equal ; 
when  A  is  negative  the  roots  are  imaginary. 

*The  sign  =  is  read  "is  identical  with,"  and  means  that  the  two  expressions 
connected  by  this  sign  differ  only  inform. 


4  ANALYTIC  GEOMETRY 

In  the  three  cases  distinguished  by  Theorem  II  the  quadratic 
may  be  written  in  three  forms  in  which  only  real  numbers  appear. 
These  are 

'  Ax2  +  Bx  -f-  C  =  A  (x—Xi)  (x  —  xz),  from  (6),  if  A  is  positive  ; 
Ax2  +  Bx  +  C  =  A(x—x1)2,  from  (6),  if  A  is  zero ; 

[-/         B  \2     ±AC-B2~] 

Ax*  +  Bx  +  C  =  A\  (*  +  o~r)  H 7-7* —  Y  if  ^  is  negative. 

L  L\        2>A)  ^A       J 

The  last  identity  is  proved  thus : 


adding  and  subtracting  —  —  -  within  the  parenthesis. 


4.  Special  quadratics.   If  one  or  both  of  the  coefficients  B  and 
C  in  (1),  p.  2,  is  zero,  the  quadratic  is  said  to  be  special. 

CASE  I.    C  =  0. 

Equation  (1)  now  becomes,  by  factoring, 

(1)  Axz  +  Bx  =  x(Ax  +  B)  =  0. 

r> 

Hence  the  roots  are  x1  =  0,  x2  =  —  —  •    Therefore  one  root  of 

yl 

a  quadratic  equation  is  zero  if  the  constant  term  of  that  equation 
is  zero.  And  conversely,  if  zero  is  a  root  of  a  quadratic,  the  con- 
stant term  must  disappear.  For  if  x  =  0  satisfies  (1),  p.  2,  by 
substitution  we  have  C  =  0. 

CASE  II.    B  =  0. 

Equation  (1),  p.  2,  now  becomes 

(2)  Ax2  +  C  =  0. 

From  Theorem  I,  p.  3,  x^  -f  x2  =  0,  that  is, 

(3)  «!=-«* 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  5 

Therefore,  if  the  coefficient  of  the  first  power  of  the  unknown 
in  a  quadratic  equation  is  zero,  the  roots  are  equal  numerically 
but  have  opposite  signs.  Conversely,  if  the  roots  of  a  quadratic 
equation  are  numerically  equal  but  opposite  in  sign,  then  the 
coefficient  of  the  first  power  of  the  unknown  must  disappear.  For, 
since  the  sum  of  the  roots  is  zero,  we  must  have,  by  Theorem  I, 
B  =  0. 

CASE  III.    B  =  C  =  0. 
Equation  (1),  p.  2,  now  becomes 
(4)  Ax2  =  0. 

Hence  the  roots  are  both  equal  to  zero,  since  this  equation 
requires  that  x2  =  0,  the  coefficient  A  beivng,  by  hypothesis, 
always  different  from  zero. 

5.  Cases  when  the  roots  of  a  quadratic  are  not  independent. 

If  a  relation  exists  between  the  roots  Xi  and  xz  of  the  Typical 

:Form  Ax9  +  Ex  +  C  =  0, 

then  this  relation  imposes  a  condition  upon  the  coefficients  A, 
B,  and  C,  which  is  expressed  by  an  equation  involving  these 
constants. 

For  example,  if  the  roots  are  equal,  that  is,  if  Xi  =  x2,  then 
B2  -4:  AC  =  0,  by  Theorem  II,  p.  3. 

Again,  if  one  root  is  zero,  then  x&2  =  0 ;  hence  C  =  0,  by 
Theorem  I,  p.  3. 

This  correspondence  may  be  stated  in  parallel  columns  thus : 

Quadratic  in  Typical  Form 

Relation  between  the  Equation  of  condition  satisfied 

roots  by  the  coefficients 

In  many  problems  the  coefficients  involve  one  or  more  arbitrary 
constants,  and  it  is  often  required  to  find  the  equation  of  condi- 
tion satisfied  by  the  latter  when  a  given  relation  exists  between 
the  roots.  Several  examples  of  this  kind  will  now  be  worked  out. 


6  ANALYTIC  GEOMETRY 

Ex.  1.    What  must  be  the  value  of  the  parameter  k  if  zero  is  a  root  of 
the  equation 
(1)  2x2-6z  +  fc2-3fc-4=0? 

Solution.    Here  A  =  2,  B  =  -  6,  C  =  k'2  -  3  k  -  4.     By  Case  I,  p.  4,  zero 
is  a  root  when,  and  only  when,  C  =  0. 


Solving,  k  =  4  or  —  1. 

Ex.  2.    For  what  values  of  k  are  the  roots  of  the  equation 

kx2  +  2kx-4x  =  2-3k 
real  and  equal  ? 

Solution.    Writing  the  equation  in  the  Typical  Form,  we  have 

(2)  fcx2  +  (2  k  -  4)x  +  (3  k  -  2)  =  0. 
Hence,  in  this  case, 

A  =  k,  B  =  2k-4,  C  =  3k-2. 
Calculating  the  discriminant  A,  we  get 

A  =  (2  k  -  4)2  -  4  k  (3  k  -  2) 
=  _8fc2-8A;-f  16  =  -8(A:2  +  &  -  2). 

By  Theorem  II,  p.  3,  the  roots  are  real  and  equal  when,  and  only  when, 

A  =  °'  .-.  k*  +  k  -  2  =  0. 

Solving,  k=  —  2  or  1.     Ans. 

Verifying  by  substituting  these  answers  in  the  given  equation  (2)  : 
when  fc=-2,  the  equation  (2)  becomes  —  2x2-8x-8=0,  or  -2(x-f  2)2=0; 
when  fc=     1,  the  equation  (2)  becomes        x2—  2  £+1=0,  or        (x—  1)2=0. 

"$Hence,  for  these  values  of  fc,  the  left-hand  member  of  (2)  may  be  trans- 
formed as  in  (7),  p.  4. 

^  Ex.  3.    What  equation  of  condition  must  be  satisfied  by  the  constants 
a^6,  fc,  and  m  if  the  roots  of  the  equation 

(3)  (62  +  a2m2)  y2  +  2a?kmy  +  a2fc2  _  a2&2  =  Q 

are  equal  ? 

Solution.    The  equation  (3)  is  already  in  the  Typical  Form  ;  hence 

A  =  62  +  a2m2,  B  =  2  a2km,  C  =  a?k2  -  a262. 
By  Theorem  II,  p.  3,  the  discriminant  A  must  vanish  ;  hence 

A  =  4  a4fc2m2  -  4  (62  +  a2m2)  (a2&2  -  a262)  =  0. 
Multiplying  out  and  reducing, 

a2&2  (fc2  -  a2m2  -  62)  =  0.     Ans. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY  7 

Ex.  4.    For  what  values  of  k  do  the  common  solutions  of  the  simultaneous 
equations 

(4)  3x  +  4y  =  fc, 

(5)  x2  +  y2  =  25 
become  identical  ? 

Solution.    Solving  (4)  for  y,  we  have 

(6)  y  =  t(*-3x), 
Substituting  in  (5)  and  arranging  in  the  Typical  Form  gives 

(7)  25  x2  -  6  kx  +~k*  -  400  =  0. 

Let  the  roots  of  (7)  be  x\  and  x2.     Then  substituting  in  (6)  will  give  the 
corresponding  values  y\  and  yz  of  y,  namely, 


(8)  yi  =  i(& 

and  we  shall  have  two  common  solutions  (xi,  y\)  and  (x2,  y2)  of  (4)  and  (5). 
But,  by  the  condition  of  the  problem,  these  solutions  must  be  identical. 
Hence  we  must  have 

(9)  xi  =  X2  and  yl  =  y2. 

If,  however,  the  first  of  these  is  true  (xi  =  x2),  then  from  (8)  y\  and  y^ 
will  also  be  equal. 

Therefore  the  two  common  solutions  of  (4)  and  (5)  become  identical  when-, 
and  only  when,  the  roots  of  the  equation  (7)  are  equal;  that  is,  when  the  dis-. 
criminant  A  of  (7)  vanishes  (Theorem  II,  p.  3). 

.-.  A  =  36  A;2  -  100  (fc*  -  400)  =  0. 
Solving,  A*  =  625, 

k  =  25  or  —  25.     Ans. 

Verification.    Substituting  each  value  of  k  in  (7), 

when  A;  =25,  the  equation  (7)  becomes  x2—  6x  +  9=0,  or  (x—  3)2=0  ;  .•.  x=3  ; 
when  k=  -25,  the  equation  (7)  becomes  x2  +  6x  +  9=0,  or  (x-f  3)2=0;  .-.x=—  3. 

Then  from  (6),  substituting  corresponding  values  of  k  and  x, 
when  k  =      25  and  x  =      3,  we  have  y  =  ^(25  —  9)=4; 
when  k  =  —  25  and  x  =  —  3,  we  have  y  =  $  (—  25  -f  9)  =  —  4. 

Therefore  the  two  common  solutions  of  (4)  and  (5)  are  identical  for  each 
of  these  values  of  fc,  namely, 

if  k  =     25,  the  common  solutions  reduce  to  x  =  3,  y  =  4  ; 
if  k  =  —  26,  the  common  solutions  reduce  to  x  =  —  3,  y  =  —  4. 

Q.E.D. 


8  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Calculate  the  discriminant  of  each  of  the  following  quadratics,  deter- 
mine the  sum,  the  product,  and  the  character  of  the  roots,  and  write  each 
quadratic  in  one  of  the  forms  (7),  p.  4. 

(a)  2x2  _  6x  +  4.  (d)  4x2  -  4x  +  1. 

(b)  x2  -  9x  -  10.  (e)  5x2  +  lOx  +  5. 

(c)  1  -  x  -  x2.  (f)  3x2  -  5x  -  22. 

2.  For  what  real  values  of  the  parameter  k  will  one  root  of  each  of  the 
following  equations  be  zero  ?. 

(a)  6x2  '+  5 kx  -  3 k2  +  3  =  0.  Ans.   k  =  ±l. 

(b)  2k  -  3x2  +  6x  -  &2  +  3  =  0.  Ans.   k  =  -  I  or  3. 

3.  For  what  real  values  of  the  parameter  are  the  roots  of  the  following 
equations  equal  ?    Verify  your  answers. 

(a)  fcx2-3x-l  =  0.  Ans.  k  =  -  f . 

(b)  x2  -  kx  +  9  =  0.  Ans.  k  =  ±  6. 

(c)  2  kx2  +  3  kx  +  12  =  0.  Ans.  k  =  ^-. 

(d)  2  x2  +  kx  -  1  =  0.  Ans.  None. 

(e)  5x2  -  3x  +  5fc2  =  0.  Ans.  k  =  ±&. 

(f)  x2  -f  kx  +  k2  +  2  =  0.  Ans.    None. 

(g)  x2  -  2 kx  -  k  -  I  =  0.  Ans.   k=-$. 

4.  Derive  the  equation  of  condition  in  order  that  the  roots  of  the  following 
equations  may  be  equal. 

(a)  m2x2  +  2  kmx  -  2px  =  -  k2.  Ans.  p  (p  -  2  km)  =  0. 

(b)  x2  +  2 mpx  +  2  6p  =  0.  Ans.  p (m2p  -  2 6)  =  0. 

(c)  2  mx2  +  2  bx  +  a2  =  0.  Ans.   fr2  =  2  a2m. 

5.  For  what  real  values  of  the  parameter  do  the  common  solutions  of  the 
following  pairs  of  simultaneous  equations  become  identical  ? 

(a)  x  +  2  y  =  k,  x2  +  y2  =  5.  -4ns.  k  =  ±  5. 

(b)  y  =  mx  —  1,  x2  =  4  y.  Ans.  m  =  ±  1. 

(c)  2 x  -  3y  =  6,  x2  +  2 x  =  3 y.  -4ns.  6  =  0. 

(d)  y  =  mx  +  10,  x2  +  y2  =  10.  .Ans.  m  =  ±  3. 

(e)  Ix  +  y  -  2  =  0,  x2  -  8  y  =  0.  Ans.  None. 

(f)  x  +  4  y  =  c,  x2  +  2  y2  =  9.  ^.ns.  c  =  ±  9. 

(g)  a2  +  y2  -  x  —  2  y  =  0,  x  +  2  y  =  c.  Ans.  c  =  0  or  5. 
(h)  x2  +  4  y2  —  8  x  =  0,  mx  -  y  -  2  m  =  0.         ^.ris.  None. 

6.  If  the  common  solutions  of  the  following  pairs  of  simultaneous  equations 
are  to  become  identical,  what  is  the  corresponding  equation  of  condition  ? 

(a)  bx  +  ay  =  a&,  y2  =  2px.  Ans.   ap  (2  62  +  ap)  =  0. 

(b)  y  =  mx  +  6,  .Ax2  +  .By  =  0. .  ^Ins.    B  (mzB  -  4  bA)  =  0. 

(c)  y  =  m(x-  a),  By2  +  Dx  =  0.  -Ans.    D(4 am2£  -  D)  =  0. 


REVIEW   OF  ALGEBRA  AND   TRIGONOMETRY 

6.  Variables.    A  variable  is  a  quantity  to  which.  jri  jjfr 
investigation,  an  unlimited  number  of  values  can  be  assigned. 
In  a  particular  problem  the  variable  may,<  in  general,  assume  any 
value  within  certain  limits  imposed  by  the  nature  of  th$  problem^ 
It  is  convenient  to  indicate  these  limits  by  inequalities. 

For  example,  if  the  variable  x  can  assume  any  value  between  —  2  and  5,  that 
is,  if  x  must  be  greater*  than  —  2  and  less  than  5,  the  simultaneous  inequalities 

*>-2,  x<5, 
are  written  in  the  more  compact  form 

-2<z<5. 

Similarly,  if  the  conditions  of  the  problem  limit  the  values  of  the  variable  x  to 
any  negative  number  less  than  or  equal'to  —  2,  and  to  any  positive  number  greater 
than  or  equal  to  5,  the  conditions 

x<—  2  OT  x  =  —  2,  and  x  >  5  or  x  =  5 
are  abbreviated  to  x  ^  —  2  and  x  ^5. 

Write  inequalities  to  express  that  the  variable 

(a)  x  has  any  value  from  0  to  5  inclusive. 

(b)  y  has  any  value  less  than  —  2  or  greater  than  —  1. 

(c)  x  has  any  value  not  less  than  —  8  nor  greater  than  2. 

7.  Equations  in  several  variables.   In  Analytic  Geometry  we 
are  concerned  chiefly  with  equations  in  two  or  more  variables. 

An  equation  is  said  to  be  satisfied  by  any  given  set  of  values 
of  the  variables  if  the  equation  reduces  to  a  numerical  equality 
when  these  values  are  substituted  for  the  variables. 

For  example,  x  =  2,  y  =  —  3  satisfy  the  equation 

2x2  +  32,2=35, 
since  2(2)2  +  3(-  3)2  =  35. 

Similarly,  x  =  —  1,  y  =  0,  z  =  —  4  satisfy  the  equation 

2z2_3y2  +  2;2_i8  =  o, 
since  2  (—  1)2  —  3  X  0  +  (—  4)2  —  18  =  0. 

*  The  meaning  of  greater  and  less  for  real  numbers  (§  1)  is  defined  as  follows  :  a  is 
greater  than  b  when  a  -  b  is  a  positive  number,  and  a  is  less  than  b  when  a  -  b  is  negative. 
Hence  any  negative  number  is  less  than  any  positive  number  ;  and  if  a  and  b  are  both 
negative,  then  a  is  greater  than  b  when  the  numerical  value  of  a  is  less  than  the  numer- 
ical value  of  6. 

Thus  3  <  5,  but  -  3  >  -  5.  Therefore  changing  signs  throughout  an  inequality  reverses 
the  inequality  sign. 


10  ANALYTIC  GEOMETRY 

An  equation  is  said  to  be  algebraic  in  any  number  of  variables, 
for  example  x,  y,  2,  if  it  can  be  transformed  into  an  equation 
each  of  whose  members  is  a  sum  of  terms  of  the  form  axmynzp, 
where  a  is  a  constant  and  m,  n,  p  are  positive  integers  or  zero. 

Thus  the  equations  x*  +  x2y2  -  z«  +  2  a:  —  5  =  0, 


are  algebraic. 

The  equation  x*  +  y*  =  a? 
is  algebraic. 

For,  squaring,  we  get    x  +  2  x$y*  +  y=a. 

Transposing,  2  x*y*  =  a  —  x  —  y. 

Squaring,  4  xy  —  a2  +  x2  +  yz  —  2  ax  —  2  ay  +  2  xy. 

Transposing,  x2  +  y2  —  2  xy  —  2  ax  —  2  ay  +  a2  =  0.                           Q.E.D. 

The  degree  of  an  algebraic  'equation  is  equal  to  the  highest 
degree  of  any  of  its  terms.*  An  algebraic  equation  is  said  to 
Ibe  arranged  with  respect  to  the  variables  when  all  its  terms  are 
transposed  to  the  left-hand  side  and  written  in  the  order  of 
descending  degrees. 

For  example,  to  arrange  the  equation 

2x'2  +  3y'  +  6x'-2x'y'-2  +  z/3  =  x'  *if  -  y'* 
with  respect  to  the  variables  z',  y',  we  transpose  and  rewrite  the  terms  in  the  order 

x'3  _  yfvtf  +  2X'2  —  2x'y'  +  y'2  +  6a/  +  3y'  —  2  =  0. 
This  equation  is  of  the  third  degree. 

An  equation  which  is  not  algebraic  is  said  to  be  transcendental. 

Examples  of  transcendental  equations  are 

y  =  sinx,  y  =  2*,  logy=3x. 

PROBLEMS 

1.  Show  that  each  of  the  following  equations  is  algebraic;  arrange  the 
terms  according  to  the  variables  x,  y,  or  x,  y,  z,  and  determine  the  degree. 


(e)  y  =  2  +  Vx2-2x-5. 
*  The  degree  of  any  term  is  the  sum  of  the  exponents  of  the  variables  in  that  term, 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY         11 


(f)  y .=  x  +  6  +  V2x2-6x 

W*=-|j> 

+  B  + 


(h)  y  =  ^ix  +  B  +  Vix'J  4-  JMTx  +  N. 

2.  Show  that  the  homogeneous  quadratic  * 

Ax2  +  Bxy  +  Cy* 

may  be  written  in  one  of  the  three  forms  below  analogous  to  (7),  p.  4,  if 
the  discriminant  A  =  B2  —  4  AC  satisfies  the  condition  given  : 

CASE     I.  Ax2  +  Bxy  +  Cy2  =  A(x-  hy]  (x  -  I2y),  if  A  >  0 ; 

CASE    II.  .Ax2  +  Bxy  +  Cy2  =  A(x-  ky)2,  if  A  =  0 ; 

',  if  A<0. 


CASE  III.  ^1x2  +  Bxy  +  Cy*  =  A  [(x  + 


8.  Functions  of  an  angle  in  a  right  triangle.  In  any  right 
triangle  one  of  whose  acute  angles  is  A,  the  functions  of  A  are 
defined  as  follows  : 


opposite  side     (/ 
sin  A  =  -**  — ,  / 

hypotenuse 

adjacent  side 
cos  A  =  -=-*  — ,    - 

hypotenuse 


esc  A  = 


sec  A 


hypotenuse     ^ 
opposite  side' 
hypotenuse 


tan  A  = 


_  opposite  side 
adjacent  side 


adjacent  side' 

adjacent  side 
cot  .4  =— J — - — ^-. 
opposite  side 


From  the  above  the  theorem  is  easily  derived : 

B  In  a  right  triangle  a  side  is  equal  to  the 
product  of  the  hypotenuse  and  the  sine  of  the 
angle  opposite  to  that  side,  or  of  the  hypote- 
nuse and  the  cosine  of  the  angle  adjacent  to 
that  side. 


A         b         C          9.  Angles    in    general.     In    Trigonometry 
an  angle  XOA   is   considered   as   gen-    < 
erated   by  the  line  OA  rotating   from 
an  initial  position  OX.     The  angle  is 
positive    when    OA    rotates    from    OX 
counter-clockwise,    and    negative    when 
the    direction    of    rotation    of    OA 
clockwise. 


s 


*The  coefficients  A,  B,  Cand  the  numbers  fj,  I2  are  supposed  real. 


12  ANALYTIC  GEOMETRY 

The  fixed  line  OX  is  called  the  initial  line,  the  line  OA  the 
terminal  line. 

Measurement  of  angles.  There  are  two  important  methods 
of  measuring  angular  magnitude,  that  is,  there  are  two  unit 
angles. 

Degree  measure.  The  unit  angle  is  ^J^  of  a  complete  revolu- 
tion, and  is  called  a  degree. 

Circular  measure.  The  unit  angle  is  an  angle  whose  subtend- 
ing arc  is  equal  to  the  radius  of  that  arc,  and  is  called  a  radian. 

The  fundamental  relation  between  the  unit  angles  is  given  by 
the  equation 

180  degrees  =  IT  radians  (IT  =  3.14159  •  •  •). 

Or  also,  by  solving  this, 

1  degree  =  ^  =  .0174  •  •  •  radians, 

loU 

1 80 
1  radian  = =  57.29  •  •  •  degrees. 

7T 

These  equations  enable  us  to  change  from  one  measurement  to 
another.  In  the  higher  mathematics  circular  measure  is  always 
used,  and  will  be  adopted  in  this  book. 

The  generating  line  is  conceived  of  as  rotating  around  0  through 
as  many  revolutions  as  we  choose.  Hence  the  important  result : 

Any  real  number  is  the  circular  measure  of  some  angle,  and 
conversely,  any  angle  is  measured  by  a  real  number. 

10.  Formulas  and  theorems  from  Trigonometry. 

1.  cotx  = ;  secx  = ;  cscx  = 

tanx  cosx  smx 

sinx  cosx 

2.  tanx  = ;  cotx  =  — — • 

cosx  sin  x 

3.  sin2x  +  cos2x  =  1 ;  1  +  tan2x  =  sec2x ;  1  +  cot2x  =  cscaz. 

4.  sin  (-  x)  =  -  sin  x ;  esc  (-  x)  =  —  esc  x  ; 
cos(—  x)  =     cosx;  sec(— x)  =     secx; 
tan(-  x)  =  — tanx ;  cot (-  x)  =  —  cotx. 


REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY         13 

5.  sin  (if  -  x)  =  sinx  ;  sin  (?r  +  x)  =  -  sinx ; 
cos  (x  -  x)  =  -  cosx  ;  cosjtf .±_&.=  — -<?osxj 
tan  (it  —  x)  =  —  tanx  jftan  (ie  +  x)  =  tanx ; 


6.  sin(-  -x)  =  cosx;  sin  (|  +  x)  =  cosx; 
cos(--x)=sinx;  cos^  +  x)  =  -  sinx; 
tan(-  -x)=cotx;  tanf-  +  x)  =  -  cotx. 


7.  sin  (2  Tt  —  x)  =  sin  (—  x)  =  —  sin  x,  etc.    O 

o 

8.  sin  (x  +  y)  =  sinx  cosy  +  cos  x  sin  y. 

9.  sin  (x  —  y)  =  sin  x  cos  y  —  cos  x  sin  y. 

10.  cos  (x  +  y}  =  cos  x  cos  y  —  sin  x  sin  y. 

11.  cos(x  —  y)  =  cos  x  cosy  +  sinx  sin  y. 

tanx  +  tan  y  .        tanx  —  tany 

12    tan  (x  +  y)  =  -  —  -  —  •        13.  tan  (x  -  y)  =  — 

I  -  tanx  tany  1  +  tanx  tany 

2  tanx 
14.  sin  2  x  =  2  sin  x  cos  x  ;  cos  2  x  =  cos2  x  —  sin2  x  ;  tan  2  x  = 


+  cosx          x 

;  tan-=  ± 

16.   Theorem.    Law  of  sines.    In  any  triangle  the  sides  are  proportional 
to  the  sines  of  the  opposite  angles; 

a  b  c 


17.  Theorem.   Lqw  of  cosines.    In  any   triangle  the  square  of  a  side 
equals  the  sum  of  the  squares  of  the  two  other  sides  diminished  by  twice  the 
product  of  those  sides  by  the  cosine  of  their  included  angle  ; 

that  is,  a2  =  62  +  c2  —  2  be  cos  A. 

18.  Theorem.    Area  of  a  triangle.    The  area  of  any  triangle  equals  one 
half  the  product  of  two  sides  by  the  sine  of  their  included  angle  ; 

that  is,  area  =  £  ab  sin  C  =  i  be  sin  A  =  ^  ca  sin  B. 


14 


ANALYTIC  GEOMETRY 


11.  Natural  values  of  trigonometric  functions. 


Angle  in 
Radians 

Angle  in 
Degrees 

Sin 

Cos 

Tan 

Cot 

.0000 

0° 

.0000 

1.0000 

.0000 

00 

90° 

1.5708 

.0873 

5° 

.0872 

.9962 

.0875 

11.430 

85° 

1.4835 

.1745 

10° 

.1736 

.9848 

.1763 

5.671 

80° 

1.3963 

.2618 

15° 

.2588 

.9659 

.2679 

3.732 

75° 

1.3090 

.3491 

20° 

.3420 

.9397 

.3640 

2.747 

70° 

1.2217 

.4363 

25° 

.4226 

.9063 

.4663 

2.145 

65° 

1.1345 

.5236 

30° 

.5000 

.8660 

.5774 

1.732 

60° 

1.0472 

.6109 

35° 

.5736 

.8192 

.7002 

1.428 

55° 

.9599 

.6981 

40° 

.6428 

.7660 

.8391 

1.192 

50° 

.8727 

.7854 

45° 

.7071 

.7071 

1.0000 

1.000 

45° 

.7854 

Cos 

Sin 

Cot 

Tan 

Angle  in 
Degrees 

Angle  in 
Radians 

Angle  in 
Radians 

Angle  in 
Degrees 

Sin 

Cos 

Tan 

Cot 

Sec 

Csc 

0 

0° 

0 

1 

0 

CO 

1 

CO 

it 
2 

90° 

1 

0 

GO 

0 

CO 

1 

it 

180° 

0 

-1 

0 

CO 

-  1 

CO 

ZTC 
2 

270° 

-1 

0 

CO 

0 

CO 

-  1 

27t 

360° 

0 

1 

0 

CO 

1 

00 

REVIEW  OF  ALGEBRA  AND  TRIGONOMETRY         15 


Angle  in 
Radians 

Angle  in 
Degrees 

Sin 

Cos 

Tan 

Cot 

Sec 

Csc 

0 

0° 

0 

1 

0 

00 

1 

CO 

it 

OQO 

1 

Vs 

V3 

-i/Q 

2  V3 

2 

6 

2 

2 

3 

Vo 

3 

it 
4 

45° 

V2 
2 

V5 

2 

1 

1 

V2 

V2 

Tt 

60° 

V3 

1 

VQ 

V3 

2 

2\/3 

3 

2 

2 

3 

3 

Tt 

2 

90° 

1 

0 

CO 

0 

«*; 

1 

12.  Rules  for  signs. 


Quadrant 

Sin 

Cos 

Tan 

Cot 

Sec 

Csc 

First   .... 

+ 

+ 

+ 

+ 

+ 

-f 

Second     .     .     . 

+ 

- 

- 

- 

- 

+ 

Third.     .     .     . 

- 

- 

+ 

+ 

- 

- 

Fourth     .     .     . 

- 

+ 

- 

- 

+ 

- 

13.  Greek  alphabet. 

Letters      Names 

Letters 

Names 

A  a         Alpha 

I    i 

Iota 

B  j8        Beta 

K    K 

Kappa 

F  y        Gamma 

A  X 

Lambda 

A  S        Delta 

MM 

Mu 

E  e         Epsilon 

N  v 

Nu 

Z  f        Zeta 

Z  $ 

Xi 

H  77        Eta 

0  o 

Omicron 

6  0        Theta 

n  TT 

Pi 

Letters 

PP 

S  <r  s 
T  r 
T  v 
*  0 
X 


Names 
Rho 
Sigma 
Tau 
Upsilon 
Phi 
Chi 
Psi 
Omega 


CHAPTER   II 
CARTESIAN   COORDINATES 

14.  Directed  line.  Let  X'X  be  an  indefinite  straight  line,  and 
let  a  point  0,  which  we  shall  call  the  origin  be  chosen  upon 
it.  Let  a  unit  of  length  be  adopted  and  assume  that  lengths 
measured  from  0  to  the  right  are  positive,  and  to  the  left  negative. 


-5-4-3-2-1    O-f-l-f-2+3  +  4+5 


Then  any  real  number  (p.  1),  if  taken  as  the  measure  of  the  length 
of  a  line  OP,  will  determine  a  point  P  on  the  line.  Conversely, 
to  each  point  P  on  the  line  will  correspond  a  real  number,  namely, 
the  measure  of  the  length  OP,  with  a  positive  or  negative  sign 
according  as  P  is  to  the  right  or  left  of  the  origin. 

The  direction  established  upon  X'X  by  passing  from  the  origin 
to  the  points  corresponding  to  the  positive  numbers  is  called  the 
positive  direction  on  the  line.  A  directed  line  is  a  straight  line  upon 


B          A          o          A         B 

which  an  origin,  a  unit  of  length,  and  a  positive  direction  have 
been  assumed. 

An  arrowhead  is  usually  placed  upon  a  directed  line  to  indicate 
the  positive  direction. 

If  A  and  B  are  any  two  points  of  a  directed  line  such  that 

OA  =  a,   OB  —  I, 

then  the  length  of  the  segment  AB  is  always  given  by  b  —  a ;  that 
is,  the  length  of  AB  is  the  difference  of  the  numbers  correspond- 
ing to  B  and  A.  This  statement  is  evidently  equivalent  to  the 
following  definition : 


CARTESIAN  COORDINATES  17 

For  all  positions  of  two  points  A  and  B  on  a  directed  line,  the 
length,  AB  is  given  by 
(1)  AB  =  OB  —  OA, 

where  0  is  the  origin. 


00  (ID  an) 

0  +3+6       —4         0+3       —30        +5       -6      -2    0 
O     A    J3       ~B        3      2*      A     0      "IT*     B       A  0 

Illustrations. 

In  Fig.  I.  AB=  OB-CU  =  6-3  =  +  3;  BA  =  OA  -  OB  =  3-6=-  3; 

II.  AB=  OB-OA=—l  —  3  =  —  7;  BA=  OA  -  OB  =  3-  (-4)  =+7; 

III.  AB  =  OB-OA  =  +  5  -(-3)=  +  8;  BA=OA-OB  =_3-5  =  -8; 

IV.  AB  =  OB-OA  =  -6-(-2)  =  -4-,  BA=OA-OB=—2-(-6)=+4. 

The  following  properties  of  lengths  on  a  directed  line  are 
obvious  : 

(2)  AB=  —  BA. 

(3)  AB  is  positive  if  the  direction  from  A  to  B  agrees  with 
the  positive  direction  on  the  line,  and  negative  if  in  the  contrary 
direction. 

The  phrase  "  distance  between  two  points  "  should  not  be  used  if  these  points 
lie  upon  a  directed  line.  Instead,  we  speak  of  the  length  AB,  remembering  that 
the  lengths  AB  and  BA  are  not  equal,  but  that  AB  =  —  BA. 

15.  Cartesian*  coordinates.   Let  X'X  and  Y'Y  be  two  directed 

lines  intersecting  at  0,  and 
^  _  oP    let  P  be  any  point  in  their 
plane.    Draw  lines  through 
P  parallel  to  X'X  and  Y'Y 
+i/  /  respectively.     Then,  if 

OM  =  a,  ON=  b, 


M 

the  numbers  a,  b  are  called 

the  Cartesian  coordinates  of 
P,  a  the  abscissa  and  b  the 
ordinate.  The  directed  lines 
X'X  and  Y'Y  are  called  the 

*  So  called  after  Ren<*  Descartes,  1596-1650,  who  first  introduced  the  idea  of  coordinates 
into  the  study  of  Geometry. 


18 


ANALYTIC  GEOMETRY 


axes  of  coordinates,  X'X  the  axis  of  abscissas,  Y'Y  the  axis  of 
ordinates,  and  their  intersection  0  the  origin. 

The  coordinates  a,  b  of  P  are  written  (a,  Z»),  and  the  symbol 
P  (a,  b)  is  to  be  read :  "  The  point  P,  whose  coordinates  are  a 
and  b." 

Any  point  P  in  the  plane  determines  two  numbers,  the  coordi- 
nates of  P.  Conversely,  given  two  real  numbers  a'  and  l>',  then 
a  point  P'  in  the  plane  may  always  be  constructed  whose  coordi- 
nates are  (a',  b').  For  lay  off  OM'  =  a',  ON'  =  b',  and  draw  lines 
parallel  to  the  axes  through  M'  and  N'.  These  lines  intersect  at 
P'(a',b').  Hence 

Every  point  determines  a  pair  of  real  numbers,  and  conversely ', 
a  pair  of  real  numbers  determines  a  point. 

The  imaginary  numbers  of  Algebra  have  no  place  in  this  repre- 
sentation, and  for  this  reason  elementary  Analytic  Geometry  is 
concerned  only  with  the  real  numbers  of  Algebra. 

16.  Rectangular  coordinates.  A  rectangular  system  of  coordi- 
nates is  determined  when  the  axes  X'X  and  Y'Y  are  perpendicular 


X' 


-4} 


to  each  other.     This  is  the  usual  case,  and  will  be  assumed  unless 
otherwise  stated. 


CARTESIAN  COORDINATES  19 

The  work  of  plotting  points  in  a  rectangular  system  is  much 
simplified  by  the  use  of  coordinate  or  plotting  paper,  constructed 
by  ruling  off  the  plane  into  equal  squares,  the  sides  being  parallel 
to  the  axes. 

In  the  figure,  p.  18,  several  points  arb  plotted,  the  unit  of  length 
being  assumed  equal  to  one  division  on  each  axis.  The  method  is 
simply  this : 

Count  off  from  0  along  X'X  a  number  of  divisions  equal  to  the 
given  abscissa,  and  then  from  the  point  so  determined  a  number 
of  divisions  equal  to  the  given  ordinate,  observing  the 
Rule  for  signs  : 

Abscissas  are  positive  or  negative  according  as  they  are  laid  off 
to  the  right  or  left  of  the  origin.     Ordinates  are 
positive  or  negative  according  as  they  are  laid 
First       off  above  or  below  the  axis  of  x. 
(+*+}   .,       Rectangular  axes  divide  the  plane  into  four 
•*  portions  called  quadrants ;  these  are  numbered 
as  in  the  figure,  in  which  the  proper  signs  of 


Second 


X'  0 


Third 


r 


Fourth 


the  coordinates  are  also  indicated. 


PROBLEMS 


1.  Plot  accurately  the  points  (3,  2),  (3,  -  2),  (-  4,  3),  (6,  0),  (-  5,  0), 
(0,  4). 

2.  Plot  accurately  the  points  (1,  6),  (3,  -  2),  (-  2,  0),  (4,  -  3),  (-  7,  -  4), 
(-  2,  4),  (0,  -  1),  ( VI,  V2),  (-  V5,  0). 

3.  What  are  the  coordinates  of  the  origin?  Ans.  (0,  0). 

4.  In  what  quadrants  do  the  following  points  lie  if  a  and  6  are  positive 
numbers:    (-a,  6)?    (-a,  -6)?    (6,  -a)?    (a,  6)? 

5.  To  what  quadrants  is  a  point  limited  if  its  abscissa  is  positive?  nega- 
tive ?  its  ordinate  is  positive  ?  negative  ? 

«-«.  Plot  the  triangle  whose  vertices  are  (2,  -  1),  (-  2,  5),  (-  8,  -  4). 

7.  Plot  the  triangle  whose  vertices  are  (—  2,  0),  (5  Vs  -  2,  5),  (—  2,  10). 

8.  Plot  the  quadrilateral  whose  vertices  are   (0,   -  2),    (4,  2),  (0,  6), 
(-4,2). 


20  ANALYTIC  GEOMETRY 

9.  If  a  point  moves  parallel  to  the  axis  of  x,  which  of  its  coordinates 
remains  constant  ?  if  parallel  to  the  axis  of  y  ? 

10.  Can  a  point  move  when  its  abscissa  is  zero?    Where?    Can  it  move 
when  its  ordinate  is  zero  ?  Where  ?     Can  it  move  if  both  abscissa  and  ordi- 
nate  are  zero  ?    Where  will  it  be  ? 

11.  Where  may  a  point  be  found  if  its  abscissa  is  2?  if  its  ordinate 
is  -3? 

12.  Where  do  all  those  points  lie  whose  abscissas  and  ordinates  are  equal  ? 

13.  Two  sides  of  a  rectangle  of  lengths  a  and  b  coincide  with  the  axes  of 
x  and  y  respectively.     What  are  the  coordinates  of  the  vertices  of  the  rec- 
tangle if  it  lies  in  the  first  quadrant  ?   in  the  second  quadrant  ?  in  the  third 
quadrant  ?  in  the  fourth  quadrant  ? 

14.  Construct  the  quadrilateral  whose  vertices  are  (-  3,  6),  (-  3,  0),  (3,  0), 
(3,  6).     What  kind  of  a  quadrilateral  is  it  ? 

15.  Join  (3,  5)  and  (-3,  -  5);  also  (3,  —  5)  and  (-  3,  5).     What  are  the 
coordinates  of  the  point  of.  intersection  of  the  two  lines  ? 


16.  Show  that  (x,  y)  and  (x,  —  y)  are  symmetrical  with  respect  to 

(x,  y)  and  (—  x,  y)  with  respect  to  Y'Y;  and  (x,  y)  and  (—  x,  —  y)  with  respect 
to  the  origin. 

17.  A  line  joining  two  points  is  bisected  at  the  origin.     If  the  coordinates 
of  one  end  are  (a,  —  6),  what  will  be  the  coordinates  of  the  other  end  ? 

18.  Consider  the  bisectors  of  the  angles  between  the  coordinate  axes. 
What  is  the  relation  between  the  abscissa  and  ordinate  of  any  point  of  the 
bisector  in  the  first  and  third  quadrants  $   second  and  fourth  quadrants  ? 

19.  A  square  whose  side  is  2  a  has  its  center  at  the  origin.     What  will  be 
the  coordinates  of  its  vertices  if  the  sides  are  parallel  to  the  axes  ?  if  the  diago- 
nals coincide  with  the  axes  ? 

Ans.   (a,  a),  (a,  -  a),  (-'a,  -  a),  (-  a,  a); 

(a  V2,  0),  (-  a  V2,  0),  (0,  a  VS),  (0,  -  a  V2). 

20.  An  equilateral  triangle  whose  side  is  a  has  its  base  on  the  axis  of  x 
and  the  opposite  vertex  above  X'X.     What  are  the  vertices  of  the  triangle  if 
the  center  of  the  base  is  at  the  origin  ?  if  the  lower  left-hand  vertex  is  at  the 
origin  ? 


CARTESIAN  COORDINATES 


21 


17.  Angles.  The  angle  between  two  intersecting  directed  lines 
is  defined  to  be  the  angle  made  by  their  positive 
directions.  In  the  figures  the  angle  between 
the  directed  lines  is  the  angle  marked  0. 

If  the  directed  lines  are  parallel,  then  the 
angle  between  them  is  zero  or  IT  according  as 
the  positive  directions  agree  or  do  not  agree. 
Evidently  the  angle  between  two  directed 
lines  may  have  any  value  from  0  to  TT  inclusive. 
Reversing  the  direction  of  either  directed  line 
changes  0  to  the  supplement  TT  —  0.  If  both  directions  are 
reversed,  the  angle  is  unchanged. 


0=0 


When  it  is  desired  to  assign  a  positive  direction  to  a  line 
intersecting  X'X,  we  shall  always  assume  the  upward  direction 
as  positive  (see  figures). 


T 

Y 

r 

^  \. 

B 

X'        0 

jS/K            y      y-'          \0 

"^^  |                    -^          -^*-                        \    /3 

r' 

x  x'         o 

\A                              Y' 

X 

A 

(1)                               (2)                                (3) 

Theorem  I.    If  a  and  ft  are  the  angles  between  a  line  directed 
upward  and  the  rectangular  axes  OX  and  OY,  then 

(I)  cos  ft  =  sin  a. 

Proof.    The  figures  are  typical  of  all  possible  cases. 


In  Fig.  1, 
and  hence 


/»=!-«, 


cos 


ft  =  cos  /  —  —  a  \  =  sin  a.    (by  6,  p.  13) 


22  ANALYTIC  GEOMETRY 

In  Fig.  2,  ft  =  a  -  1 9 

I  -A 

and  hence  cos  J3  =  cos  (  a  —  -- 1  =  sin  a. 

(by  4  and  6,  p.  12) 

In  Fig.  3,  a  =  ^,  ft  =  0. 

.'.  cos  ft  =  1  =  sin  a.  Q.E.D. 

The  positive  direction  of  a  line  parallel  to  X'X  will  be  assumed 
to  agree  with  the  positive  direction  of  X'X,  that  is,  to  the  right. 

Hence  for  such  a  line  a  =  0,  ft  =  —  >  and  the  relation  (I)  still 
holds,  since 

7T 

cos  8  =  cos  —  =  0  =  sin  0  =  sin  a.  * 


PROBLEMS 

1.  Show  that  for  lines  directed  downward  cos  /3  =  —  sin  a. 

2.  What  are  the  values  of  a  and  0  for  a  line  directed  N.E.  ?  N.  W.  ?   S.E.  ? 
S.W.  ?     (The  axes  are  assumed  to -indicate  the  four  cardinal  points  of  the 
compass.) 

3.  Find  the  relation  between  the  a's  and  /3's  of  two  perpendicular  lines 
directed  upward.  Ans    a'_a  =  5;    p  +  p  =  ?L. 

2  2 

18.  Orthogonal  projection.    The  orthogonal  projection  of  a  point 
upon  a  line  is  the  foot  of  the  perpendicular 
let  fall  from  the  point  upon  the  line. 

Thus  in  the  figure 
M  is  the  orthogonal  projection  of  P  on  X'X',         M 


N  is  the  orthogonal  projection  of  P  on  FT;  X'  J/  - 

Pf  is  the  orthogonal  projection  of  P' on  X'X.          ° — 

If  A  and  B  are  two  points  of  a  directed 
line,  and  7kf  and  N  their  projections  upon  a 
second  directed  line  CD,  then  MN  is  called  the  .projection  of  AB 
upon  <7Z>. 


CARTESIAN  COORDINATES 


23 


Theorem  II.    First  theorem  of  projection.    If  A  and  B  are  points 
upon  a  directed  line  making  an  angle  y  with  a  second  directed  line 
CD,  then  the 
(II)         projection  of  the  length  AB  upon  CD  =  AB  cos  y. 

Proof.    In  the  figures  let 

a  =  the  numerical  length  of  AB, 
I  ±=  the  numerical  length  of  AS  or  BT\ 

then  a  and  I  are  positive  numbers  giving  the  lengths  of  the  respec- 
tive lines,  as  in  Plane  Geometry.  Now  apply  the  definition  of  the 
cosine  to  the  right  triangles  ABS  and  ABT  (p.  11). 

A  A    B  -      «T      A  >    B     A  >    B 

rNs»         N    _^\M  $\~}MD   ! 

C\M  BK^JTIX^    !    c   ^CT     i    .   i     i    t  ! 
"i>  5  r  M     N  M     N 

(1)  (2)  (3)  (4)  (5)  C6) 


[M 


In  Fig.  1, 


In  Fig.  2, 


In  Fig.  3, 


In  Fig.  4, 


In  Fig.  5, 
Hence 

In  Fig.  6, 
Hence 


I  =  a  cos  BA  S  =  a  cos  y, 


'.  MN  —  AB  COS  y. 

£  =  <z  cos  ABT  =  a  cos  (TT  —  y) 

=  —  a  cos  y,  (by  5,  p.  13) 

MN  =1,  AB=—a. 


I  —  a  cos  ABT  =  a  cos  (TT  —  y) 

=  —  a  cos  y, 
MN  =  —  J,  ^^  =  a. 
.'.  3/^  =  ^4.6  cos  y. 

Z  =  a  cos  ^.BT7  =  a  cos  y, 
Jlf  JV  =  -  19  AB  =  -a. 
.'.  MN  =  AB  cos  y. 

y  =  0,  MN  =  I,  AB  =  a. 
MN  =  AB  =  AB  cos  0  (since  cos  0  =  IV 
.*.  MN  =  ^45  cos  y. 

y  =  TT,  J/iY  =  —  /,  ^4j5  =  a. 
Jlf^V  =  —  AB  =  ABcosjr  (since  cos  TT  =  —  1). 
y.  Q>E> 


24  ANALYTIC  GEOMETRY 

19.  Lengths.    Consider  any  two  given  points 

Then  in  the  figure 

MiMt  =  projection  of  PiP2  on  X'X, 
NiNz  =  projection  of  PiPa  on  YY. 

Y 
#i 


Mz 


X' 


But  by  (1),  p.  17, 


o 


Mi  . 


** 


(III) 


Hence 

Theorem  HI.    Given  any  two  points  Pl  (cc1?  y^),  P2  (#2,  y2) ; 
2  —  &!  =  projection  of  JP^ Ft  on 
!  —  1/1  =  projection  of  P^fz  on 

We  may  now  easily  prove  the  important 
Theorem  IV.    The  length  I  of  the  line  Y' 

joining  two  points  P!  (a^,  3^),  P2  (x2,  y2)  ^  2 

is  given  by  the  formula  ^ 

(IV)  1  = 


Proof.    Draw  lines  through  P!  and  x> 
PZ  parallel  to  the  axes  to  form  the 
right  triangle  PiSP2.  r 

Then  SP,  = 


(x\,yi 


« 


t  x 


PzS  = 


yi- 


and  hence 


(by  III) 
(by  III) 


Q.E.D. 


CARTESIAN  COORDINATES 


25 


The  method  used  in  deriving  (IV)  for  any  positions  of  PI  and 
P2  is  the  following  : 

Construct  a  right  triangle  by  drawing  lines  parallel  to  the  axes 
through  P!  and  P2.  The  sides  of  this  triangle  are  equal  to  the 
projections  of  the  length  P^PZ  upon  the  axes.  But  these  projec- 
tions are  always  given  by  (HI),  or  by  (III)  with  one  or  both 
signs  changed.  The  required  length  is  then  the  square  root  of 
the  sum  of  the  squares  of  these  projections,  so  that  the  change  in 
sign  mentioned  may  be  neglected.  A  number  of  different  figures 
should  be  drawn  to  make  the  method  clear. 

Ex.  1.    Find  the  length  of  the  line  joining  the  points  (1,  3)  and  (—  5,  6). 

Solution.    Call  (1,  3)  Plt  and  (-  6,  5)P2. 
Then 

«i  =  1,  y\  =  3,  and  Xa  =  -  5,  y2  =  5 ; 
and  substituting  in  (IV),  we  have 
I  =  V(l  +  5)2  +  (3  -  5)2  =  V40  =  2  VlO. 


It  should  be  noticed  that  we  are  simply 
finding  the  hypotenuse  of  a  right  triangle 
whose  sides  are  6  and  2. 


Remark.  The  fact  that  formulas  (III)  and  (IV)  are  true  for  all 
positions  of  the  points  Pl  and  P2  is  of  fundamental  importance. 
The  application  of  these  formulas  to  any  given  problem  is  there- 
fore simply  a  matter  of  direct  substitution,  as  the  example  worked 
out  above  illustrates.  In  deriving  such  general  formulas,  since 
it  is  immaterial  in  what  quadrants  the  assumed  points  lie,  it  is 
most  convenient  to  draw  the  figure  so  that  the  points  lie  in  the 
first  quadrant,  or,  in  general,  so  that  all  the  quantities  assumed 
as  known  shall  be  positive. 

PROBLEMS 

1.  Find  the  projections  on  the  axes  and  the  length  of  the  lines  joining 
the  following  points : 

(a)  (-4,  -  4)  and  (1,  3).  Ans.  Projections  6,  7;  length  =  V74. 

(b)  (- V2,  V3)  and  (\/3,  V2). 

-4ns.  Projections  Vjj  +  \/2,  V2  —  V§ ;  length  =  VlO. 


26  ANALYTIC  GEOMETRY 


(c)  (0,  0)  and  (-,  -   -V  Ans.  Projections  -,  -  V3;  length  =  a. 

\2        2    /  22 


(d)  (a  +  6,  c  +  a)  and  (c  +  a,  6  +  c). 


Ans.   Projections  c  -  6,  b  -  a;  length  =  v(6  -  c)2+(a  -  b)2. 

2.  Find  the  projections  of  the  sides  of  the  following  triangles  upon  the 

axes: 

(a)  (0,  6),  (1,  2),  (3,  -  5). 

(b)  (1,  0),  (-  1,  -  5),  (-  1,  -  8). 

(c)  (a,  6),  (6,  c),  (c,  d). 

3.  Find  the  letfgths  of  the  sides  of  the  triangles  in  problem  2. 

4.  Work  out  formulas  (III)  and  (IV),  (a)  if  Xi  =  xz ;  (b)  if  y\  —  y2. 

5.  Find  the  lengths  of  the  sides  of  the  triangle  whose  vertices  are  (4,  3), 
(2,  -2),  (-3,  5). 

6.  Show  that  the  points  (1,  4),  (4, 1),  (5,  5)  are  the  vertices  of  an  isosceles 
triangle. 

7.  Show  that  the  points  (2,  2),  (-  2,  -  2),  (2  V3,  -  2  Vjj)  are  the  vertices 
of  an  equilateral  triangle. 

8.  Show  that  (3,  0),  (6,  4),  (—  1,  3)  are  the  vertices  of  a  right  triangle. 
What  is  its  area  ? 

9.  Prove  that  (-  4,  —  2),  (2,  0),  (8,  6),  (2,  4)  are  the  vertices  of  a 
lelogram.     Also  find  the  lengths  of  the  diagonals. 

10.  Show  that  (11,  2),  (6,  -  10),  (-6,  -  5),  (- 1,  7)  are 
a  square.     Find  its  area. 

11.  Show  that  .the  points  (1,  3),  (2,  Ve),  (2,  -  V6) 

the  origin,  that  is,  show  that  they  lie  on  a  circle  with  its  cen 
and  its  radius  VlO. 

12.  Show  that  the  diagonals  of  any  rectangle  are  equal.  v 

13.  Find  the  perimeter,  of  the  triangle  whose  vertices  are  (a,  6),  (—  a,  6), 
(-a,  -6). 

14.  Find  the  perimeter  of  the  polygon  formed  by  joining  the  following 
points  two  by  two  in  order : 

(6,  4),  (4,  -  3),  (0,  - 1),  (-  5,  -  4),  (-  2,  1). 

15.  One  end  of  a  line  whose  length  is  13  is  the  point  (-  4,  8);  the  ordi- 
nate  of  the  other  end  is  3.     What  is  its  abscissa  ?  Ans.  8  or  -  16. 

16.  What  equation  must  the  coordinates  of  the  point  (x,  y)  satisfy  if  its 
distance  from  the  point  (7,  -  2)  is  equal  to  11  ? 


CARTESIAN  COORDINATES 


27 


17.  What  equation  expresses  algebraically  the  fact  that  the  point  (z,  y)  is 
equidistant  from  the  points  (2,  3)  and  (4,  5)? 

^—18.  If  the  angle  XOY  (Fig.,  p.  17)  equals  o>,  show  that  the  length  of  tha 
line  joining  PI(ZI,  yi)  and  P2  (x2,  ?/2)  is  given  by 

I  =  V(xi  -  z2)2  +  (yl  -  y2)2  +  2  (X!  -  x2)  (yi  -  y2)  cos  w, 

"^19.  If  w  =  -,  find  distance  between  the  points  (—  3,  3)  and  (4,  —  2). 

Ans.  V39. 

20.  If  w  =  — ,  find  the  perimeter  of  the  triangle  whose  vertices  are  (1,  3), 
3 


(2,  7),  (-4,  -4). 


Ans. 


21.  If  w  =  -,  find  the  perimeter  of  triangle  (1,  2),  (-  2,  -  4),  (3,  -  5). 
6 


Ans.  3  Vs  +  2  V3  +  V26  -  5  V3  +  Vs3  -  14  Vg. 

22.  Prove  that  (6,  6),  (7,  —  1),  (0,  -2),  (—2,  2)  lie  on  a  circle  whose 
center  is  at  (3,  2). 

23.  If  w  =  — ,  find  the  distance  between  (\/3,  -x/2),  (-  \/2,  V§). 

Ans.   VlO  +  V2. 

24.  Show  that  the  sum  of  the  projections  of  the  sides  of  a  polygon  upon 
eitlfll  axis  is  zero  if  each  side  is  given  a  direction  established  by  passing 

luously  around  the  perimeter. 

ion  and  slope.   The  inclination  of  a  line  is  the  angle 
is  of  x  and  the  line  when  the  latter  is  given  the 
upward  direction  (p.  21). 

The  slope  of  a  line  is  the  tangent  of 
its  inclination. 

The  inclination  of   a  line  will  be 
denoted  by  a,  a^  azj  a',  etc. ;  its  slope 
>  by  w,  m1?  m2,  m',  etc.,  so  that  m  =  tan  a, 
ml  —  tan  al9  etc. 

The  inclination  may  be  any  angle 
from  0  to  TT  inclusive  (p.  21).  The 
slope  may  be  any  real  number,  since  the  tangent  of  an  angle  in 
the  first  two  quadrants  may  be  any  number  positive  or  negative. 
The  slope  of  a  line  parallel  to  X'X  is  of  course  zero,  since  the 
inclination  is  0  or  TT.  For  a  line  parallel  to  Y'  Y  the  slope  is  infinite. 


28 


ANALYTIC  GEOMETRY 


00 


Theorem  V.    The  slope  m  of  the  line  passing  through  two  points 
(xii  yi)>  PI  (xiu  2/2)  **  given  by 

Vi  —  Vi 


Proof. 


(1) 


Similarly, 


(2) 


=  x2  —  xl 
=  PiP2  cos  or. 

cos  a  =  #2  —  Xi. 

&!&*  =  ^2  —  2/i 

=  PjP2  COS  /3. 

COS  /3  =  2/2  —  2/1- 
cos  /?  =  sin  a. 


But 

Hence,  from  (2), 

(3)  PiPa  sin  a  =  2/2  -  2/i- 

Dividing  (3)  by  (1),     tan  a  =  m  = 

g  ~~      ! 

Remark.  Formula  (V)  may  be  verified  by 
constructing  a  right  triangle  whose  hypot- 
enuse is  PiP2,  as  on  p.  24,  whence  tan  a 
(=  tan  Z  SPiP2)  is  found  directly  as  the  ratio 
of  the  opposite  side,  SP2  =  yz  —  yw  to  the 
adjacent  side,  PjS  =  x2  —  xlt* 


(by  (III),  p.  24) 
(by  (II),  p.  23) 

(by  (III),  p.  24) 
(by  (II),  p.  23) 


(by  (I),  p.  21) 


Q.E.D. 


*  To  construct  a  line  passing  through  a  given  point  P^  whose  slope  is  a  positive  frac- 
tion ^ ,  we  mark  a  point  S  b  units  to  the  right  of  P^  and  a  point  P,  a  units  above  S,  and 
draw  P,P2.  If  the  slope  is  a  negative  fraction,  —  | ,  then  either  S  must  lie  to  the  left  of  Pt 
or  P,  must  lie  below  S. 


CARTESIAN  COORDINATES  29 

Theorem  VI.  If  two  lines  are  parallel,  their  slopes  are  equal;  if 
perpendicular,  the  slope  of  one  is  the  negative  reciprocal  of  the  slope 
of  the  other,  and  conversely. 

Proof.  Let  a-^  and  <x2  be  the  inclinations  and  m^  and  m2  the 
slopes  of  the  lines. 

If  the  lines  are  parallel,  <xl  =  a2.     .'.  m^  —  m2. 

If  the  lines  are  perpendicular,  as  in  the  figure, 


,'.  m1  =  tan  a^  =  tan  I  az  —  —  \ 


=  —  cot  a2  (by  4  and  6,  p.  12) 

1 


V 


Q.E.D. 


The  converse  is  proved  by  retracing  the  steps  with  the  assump- 
tion, in  the  second  part,  that  a2  is  greater  than  av 


PROBLEMS 


.  Find  the  slope  of  the  line  joining  (1,  3)  and  (2,  7).  Ans.   4. 

\y:Fmd  the  slope  of  the  line  joining  (2,  7)  and  (—  4,  —  4).       Ans.  *£-. 
jl. 


Find  the  slope  of  the  line  joining  (  V§,  V2)  and  (-  V2,  V3). 
/  Ans.  2V 

Find  the  slope  of  the  line  joining  (a  +i,  c  +  a),  (c  +  a,  b  +  c). 


5.  Find  the  slopes  of  the  sides  of  the  triangle  whose  vertices  are  (1,  1), 

(-  1,  -  1),  (  V3,  -  V3).  j/  1+V3    1-V31 

-d.rw."i,  -  —  ,  -  —  r 

1_V§    1+V3 

6.  Prove  by  means  of  slopes  that  (-  4,  —  2),  (2,  0),  (8,  6),  (2,  4)  are  the 

vertices  of  a  parallelogram.      I/O     % 
/  *>  2 

/  7.  Prove  by  means  of  slopes  that  (3,  0),  (6,  4),  (-  1,  3)  are  the  vertices 

of  a  right  triangle. 

8.  Prove  by  means  of  slopes  that  (0,  -2),  (4,  2),  (0,  6),  (-4,  2)  are  the 
vertices  of  a  rectangle,  and  hence,  by  (IV),  of  a  square. 


".  . 


30  ANALYTIC  GEOMETRY 

9.  Prove  by  means  of  their  slopes  that  the  diagonals  of  the  square  in 
problem  8  are  perpendicular. 

10.  Prove  by  means  of  slopes  that  (10,  0),  (5,  5),  (5,  —  5),  (-  5,  5)  are 
the  vertices  of  a  trapezoid. 

11.  Show  that  the  line  joining  (a,  6)  and  (c,  —  d)  is  parallel  to  the  line 
joining  (—a,  —  6)  and  (—  c,  d). 

12.  Show  that  the  line  joining  the  origin  to  (a,  6)  is  perpendicular  to  the 
line  joining  the  origin  to  (—6,  a). 

13.  What  is  the  inclination  of  a  line  parallel  to  Y'Y?    perpendicular  to 

Y'Y? 

14.  What" is  the  slope  of  a  line  parallel  to  Y'Y?    perpendicular  to  Y'Y? 


15.  What  is  the  inclination  of  the  line  joining  (2,  2)  and  (-  2,  -  2)? 

Ans.  • 

16.  What  is  the  inclination  of  the  line  joining  (—  2,  0)  and  (-  5,  3)? 

Ans. 

17.  What  is  the  inclination  of  the  line  joining  (3,  0)  and  (4,  V3)  ? 


Ans.  —  • 


Ans.  f- 

18.  What  is  the  inclination  of  the  line  joining  (3,  0)  and  (2,  V3)  ? 

Ans.  ^- 

19.  What  is  the  inclination  of  the  line  joining  (0,  -  4)  and  (-  V§,  -  5)  ? 

Ans.  —  - 

20.  What  is  the  inclination  of  the  line  joining  (0,  0)  and  (-  V§,  1)  ? 

5* 
Ans.  —  • 

21.  Prove  by  means  of  slopes  that  (2,  3),  (1,  -  3),  (3,  9)  lie  on  the  same 
straight  line. 

22.  Prove  that  the  points  (a,  b  +  c),  (6,  c  +  a),  and  (c,  a  +  b)  lie  on  the 
same  straight  line. 

23.  Prove  that  (1,  5)  is  on  the  line  joining  the  points  (0,  2)  and  (2,  8) 
and  is  equidistant  from  them. 

24.  Prove  that  the  line  joining  (3,  —  2)  and  (5,  1)  is  perpendicular  to  the 
line  joining  (10,  0)  and  (13,  -  2). 


CARTESIAN   COORDINATES  31 

21.  Point  of  division.    Let  Px  and  P2  be  two  fixed  points  on  a 
directed  line.     Any  third  point  on  the  line,  as  P  or  P',  is  said 


.PI  P       P2        P' 

"to  divide  the  line  into  two  segments,"  and  is  called  a  point 
of  division.  The  division  is  called  internal  or  external  according 
as  the  point  falls  within  or  without  PiP2.  The  position  of  the 
point  of  division  depends  upon  the  ratio  of  its  distances  from  Pj 
and  P2.  Since,  however,  the  line  is  directed,  some  convention 
must  be  made  as  to  the  manner  of  reading  these  distances.  We 
therefore  adopt  the  rule : 

If  P  is  a  point  of  division  on  a  directed  line  passing  through 
P!  and  P2,  then  P  is  said  to  divide  PiP2  into  the  segments  PXP 

P  P 
and  PP2.     The  ratio  of  division  is  the  value  of  the  ratio*  — - — 

We  shall  denote  this  ratio  by  X}  that  is, 


If  the  division  is  internal,  PXP  and  PP2  agree  in  direction  and 
therefore  in  sign,  and  A.  is  therefore  positive.  In  external  divi- 
sion X  is  negative.  ,  The  sign  of  X  therefore  indicates  whether 
the  point  of  division  P  is  within  or  without  the  segment  PiP2; 
and  the  numerical  value  determines  whether  P  lies  nearer  P^ 
or  P2.  The  distribution  of  X  is  indicated  in  the  figure. 

&       •     •' ' 

-1<\<0  Ac=0        \>0         X  =  oo      -oo<X<-l 

-PI          "V 

That  is,  X  may  have  any  positive  value  between  Px  and  P2,  any 
negative  value  between  0  and  —  1  to  the  left  of  Plt  and  any  nega- 
tive value  between  —  1  and  —  oo  to  the  right  of  P2.  The  value 
—  1  for  X  is  excluded. 

*  To  assist  the  memory  in  writing  down  this  ratio,  notice  that  the  point  of  division  P 
is  written  last  in  the  numerator  and^rsi  in  the  denominator. 


32 


ANALYTIC  GEOMETRY 


Introducing  coordinates,  we  next  prove 

Theorem  VII.  Point  of  division.  The  coordinates  (x,  y)  of  the 
point  of  division  P  on  the  line  joining  PI(#I,  T/J),  P2(#2,  y2),  such 
that  the  ratio  of  the  segments  is 


are  given  by  the  formulas 

*   j_  Jl 

(VII) 


0?  = 


'9  y  = 


Proof.    Given 


PIP 
pp2 


/  0 


Ml    M 


Let  a  be  the  inclination  of  the  line  PiP2.     Project  P1?  P,  P2 
upon  the  axis  of  x. 

Then,  by  the  first  theorem  of  projection  [(II),  p.  23], 


M^M  =  PjP  cos  ocj 
MM2  =  PP2  cos  a. 


Dividing, 
But 

Substituting, 

Clearing  of  fractions  and  solving  for  x, 

«,  -f  AiC2 

*=iTr 


MM2      PP2 

MM2  =  x2  —  x. 
~ i \ 

—  A. 


(by  hypothesis) 
(by  (III),  p.  24) 


Similarly, 


y 


y\  +  Xy2 

1+A 


Q.E.D. 


Corollary.  Middle  point.  The  coordinates  (x,  y)  of  the  middle 
point  of  the  line  joining  P1  (x^  y^),  P2  (ic2,  y2)  a™  found  by  taking 
the  averages  of  the  given  abscissas  and  ordinates;  that  is, 


For  if  P  is  the  middle  point  of  P^,  then  A  =         =  1. 

PP2 


CARTESIAN  COORDINATES 


33 


Ex.  1.    Find  the  point  P  dividing  PI  ( - 1,  - 6),  P2  (3,  0)  in  the  ratio  X  =  -  J. 
Solution.    Applying  (VII),  Xi  =  -  1,  2/1  =  -  6, 


a 

.0, 

o 

/ 

I'l 

/^ 

/ 

/ 

/ 

4 

(J 

-6> 

/ 

yl 

/ 

f 

P 

f 
T 

-4ns. 


Hence  Pis  (-  2£,  -  8). 

Ex.  2.  Find  the  coordinates  of  the  point  of 
intersection  of  the  medians  of  a  triangle  whose 
vertices  are  (xl5  ?/i),  (x2,  y2),  (xs,  ys). 

Solution.  By  Plane  Geometry  we  have  to  find 
the  point  P  on  the  median  AD  such  that  AP  =  f  AD,  that  is,  AP :  PD : :  2  : 1, 
or  \  =  2. 

By  the  Corollary,  D  is  [i  (x2  +  x3),  \  (y2  +  ys)].  «^f  *j»yJ 

To  find  P,  apply  (VII),  remembering  that  A  corre- 
sponds to  (xi,  y\)  and  D  to  (x2,  y2). 


This  gives 


x  = 


y  = 


1  +  2 


1  +  2 
x  = 


x2  +  X3),  y  =  i  (2/1  +  yz  +  2/s)- 

Hence  the  abscissa  of  the  intersection  of  the  medians  of  a  triangle  is  the 
average  of  the  abscissas  of  the  vertices,  and  similarly  for  the  ordinate. 

The  symmetry  of  these  answers  is  evidence  that  the  particular  median 
chosen  is  immaterial,  and  the  formulas  therefore  prove  the  fact  of  the  intersec- 
tion of  the  medians. 


1 

and 

2 

and 

3 


d 

and 

5 


PROBLEMS 

Find  the  coordinates  of  the  middle  point  of  the  line  joining  (4,  —  6) 
(-  2,  -  4).  Ans.    (1,  -  5). 

Find  the  coordinates  of  the  middle  point  of  the  line  joining  (a +  6,  c  +  d) 
(a  -  6,  d  —  c).  Ans*  (a,  d). 

Find  the  middle  points  of  the  sides  of  the  triangle  whose  vertices  ai 
),  (4,  -  6),  and  (-3,  —  6) ;  also  find  the  lengths  of  the  medians. 

Find  the  coordinates  of  the  point  which  divides  the  line  joining  (-1,4) 
(-5,  -  8)  in  the  ratio  1:3.  .  Ans.    (-  2,  1). 

Find  the  coordinates  of  the  point  which  divides  the  line  joining 
!,  -  5)  and  (6,  9)  in  the  ratio  2  : 6.  Ana.    (-  $,  - 1). 


34  ANALYTIC  GEOMETRY 

6.  Find  the  coordinates  of  the  point  which  divides  the  line  joining  (2,  6) 
and  (—  4,  8)  into  segments  whose  ratio  is  —  f .  Ans.    (  —  22,  14). 

7.  Find  the  coordinates  of  the  point  which  divides  the  line  joining 
(—3,  —  4)  and  (5,  2)  into  segments  whose  ratio  is  —  f.   Ans.  (—19,  —  16). 

8.  Find  the  coordinates  of  the  points  which  trisect  the  line  joining  the 
points  (-  2,  -  1)  and  (3,  2).  Ans.    (-  |,  0),  (f,  1). 

9.  Prove  that  the  middle  point  of  the  hypotenuse  of  a  right  triangle  is 
equidistant  from  the  three  vertices. 

10.  Show  that  the  diagonals  of  the  parallelogram  whose  vertices  are  (1,2), 
(-  5,  -  3),  (7,  -  6),  (1,  -  11)  bisect  each  other. 

11.  Prove  that  the  diagonals  of  any  parallelogram  mutually  bisect  each 
other. 

12.  Show  that  the  lines  joining  the  middle  points  of  the  opposite  sides  of 
the  quadrilateral  whose  vertices  are  (6,  8),  (—  4,  0),  (—  2,  —  6),  (4,  —  4)  bisect 
each  other. 

13.  In  the  quadrilateral  of  problem  12  show  by  means  of  slopes  that  the 
lines  joining  the  middle  points  of  the  adjacent  sides  form  a  parallelogram. 

14.  Show  that  in  the  trapezoid  whose  vertices  are  (—  8,  0),  (—4,  —  4), 
(—  4,  4),  and  (4,  —  4)  the  length  of  the  line  joining  the  middle  points  of  the 
non-parallel  sides  is  equal  to  one  half  the  sum  of  the  lengths  of  the  parallel 
sides.     Also  prove  that  it  is  parallel  to  the  parallel  sides. 

-     15.  In  what  ratio  does  the  point  (—  2,  3)  divide  the  line  joining  the  points 
(-3,  5)  and  (4,  -9)?  Ans.    1. 

16.  In  what  ratio  does  the  point  (16,  3)  divide  the  line  joining  the  points 
(-5,0)  and  (2,  1)?  Ans.    -  f. 

17.  Given  the  triangle  whose  vertices  are  (—  5,  3),  (1,  —  3),  (7,  5);  show 
that  a  line  joining  the  middle  points  of  any  two  sides  is  parallel  to  the  third 
side  and  equal  to  one  half  of  it. 

18.  If  (2,  1),  (3,  3),  (6,  2)  are  the  middle  points  of  the  sides  of  a  triangle, 
what  are  the  coordinates  of  the  vertices  of  the  triangle  ? 

Ans.    (-1,2),  (5,0),  (7,4). 

19.  Three  vertices  of  a  parallelogram  are  (1,  2),  (-5,  —3),  (7,  -6). 
What  are  the  coordinates  of  the  fourth  vertex  ? 

Ans.    (1,  -  11),  (-  11,  5),  or  (13,  -  1). 

D.'  The  middle  point  of  a  line  is  (6,  4),  and  one  end  of  the  line  is  (5,  7). 
What  are  the  coordinates  of  the  other  end  ?  Ans.    (7,  1). 

>    21.  The  vertices  of  a  triangle  are  (2,  3),  (4,  -  5),  (-  3,  -  6).     Find  the  4 
coordinates  of  the  point  where  the  medians  intersect  (center  of  gravity). 


CARTESIAN  COORDINATES 


35 


22.  Find  the  area  of  the  isosceles  triangle  whose  vertices  are  (1,  5),  (5,  1), 
(_  o?  _  9)  by  finding  the  lengths  of  the  base  and  altitude. 

23.  A  line  AB  is  produced  to  C  so  that  BC  =  \  AB.    If  the  points  A  and  B 
have  the  coordinates  (5,  6)  and  (7,  2)  respectively,  what  are  the  coordinates 
of  C?  Ans.    (8,  0). 

24.  Show  that  formula  (VII)  holds  for  oblique  coordinates,  that  is,  Z  XOT 
may-  have  any  value. 

25.  How  far  is  the  point  bisecting  the  line  joining  the  points  (5,  5)  and  (3,  7) 
from  the  origin  ?     What  is  the  slope  of  this  last  line  ?          Ans.   2  Vl3,  f  . 

22.  Areas.  In  this  section  the  problem  of  determining  the  area 
of  any  polygon  the  coordinates  of  whose  vertices  are  given  will 
be  solved.  We  begin  with 

Theorem  VIII.  The  area  of  a  triangle  whose  vertices  are  the 
origin,  PI(XI}  y-^),  and  P2(#2,  y?)  is  given  by  the  formula 

(VIII)         Area  of  triangle  OPiP2  =  |(xl2/2  -  a?22/1). 

Proof.    In  the  figure  let 


=  Z.  XOP2, 


(1) 


By  18,  p.  13, 
(2)  Area  A 


1.  OP2  sin  0 


(3) 


But  in  the  figure 


OPl  .  OP2  (sin  ft  cos  a  -  cos  /3  sin  a). 

(by  9,  p.  13) 


sin  a  = 


OP2        OP2 


OP1 


OP2       OP2 


, 

cos  a  =  --  -1 


OPl       OPl 


Substituting  in  (3)  and  reducing,  we  obtain 
Area  A  OP^  =  %  (Xly2  -  x&J. 


Q.E.D. 


36 


ANALYTIC  GEOMETRY 


Ex.  1.    Find  the  area  of  the  triangle  whose  vertices  are  the  origin,  (—2,  4), 
and  (-5,  -1). 

Solution.    Denote  ( -  2, 4)  by  PI,  ( -  5,  - 1)  by  P2. 
Then 

Substituting  in  (VIII), 


Then  Area  =  11  unit  squares. 

If,  however,  the  formula  (VIII)  is  applied  by  denoting  (—  2,  4)  by  P2,  and 
(-5,  -  1)  by  PI,  the  result  will  be  -  11. 
The  two  figures  are  as  follows : 


(1)  (2) 

The  cases  of  positive  and  negative  area  are  distinguished  by 
Theorem  IX.    Passing  around  the  perimeter  in  the  order  of  the 
vertices  0,  P19  P2, 

if  the  area  is  on  the  left,  as  in  Fig.  1,  then  (VIII)  gives  a  posi- 
tive result; 

if  the  area  is  on  the  right,  as  in  Fig.  2,  then  (VIII)  gives  a 
negative  result. 

Proof.    In  the  formula 

(4)  Area  A  OP^  =  ±OPl-  OP2  sin  $ 

the  angle  6  is  measured  from  OPl  to  OP2  within  the  triangle. 
Hence  0  is  positive  when  the  area  lies  P 
to  the  left  in  passing  around  the  per- 
imeter  0,  P19  P2,  as  in  Fig.  1,  since  0  is  V 
then  measured  counter-clockwise  (p.  11). 
But  in  Fig.  2,  0  is  measured  clockwise. 
Hence  0  is  negative  and  sin  0  in  (4)  is  also  negative.  Q.E.D. 

Formula  (VIII)  is  easily  applied  to  any  polygon  by  regarding 
its  area  as  made  up  of  triangles  with  the  origin  as  a  common 
vertex.  Consider  any  triangle. 


(2) 


CARTESIAN  COORDINATES  37 

Theorem  X.    The  area  of  a  triangle  whose  vertices  are  PI(XI}  y^, 
-Pa(«»  2/2),  ^a  (*3,  2/a)  w  given  by 
(X)    Area  A  P^P^P^ = \  (a^y,  - #22/i + #22/3  - a?3y2 + xzyv  - o^y,). 

This  formula  gives  a  positive  or  negative  result  according  as  the 
area  lies  to  the  left  or  right  in  passing 
around  the  perimeter  in  the  order  P^P^P^. 

Proof.    Two   cases   must   be   distin- 
^      guished    according    as    the    origin    is 
within  or  without  the  triangle. 

Fig.   1,    origin   within   the   triangle. 
By  inspection, 

(5)  Area  A  Pff^  =  A  OP^  +  A  OP2P3  +  A  OPfv 
since  these  areas  all  have  the  same  sign. 

Fig.  2,  origin  without  the  triangle.     By  inspection, 

(6)  Area  A  P,P2P3  =  A  OPJ>%  +  A  OP2P3  +  A  OP3P» 

since  OP^P^  OPZP1  have  the  same  sign,  but  OP2P3  the  opposite 
sign,  the  algebraic  sum  giving  the  desired  area. 

By  (VIII),  A  OP,P2  =  ±(XM  -  x&J, 

A  OP2P3  =  J  (^27/3  -  x3y2),  A  OPgPi  =  i  (a?^  -  «,y8). 

Substituting  in  (5)  and  (6),  we  have  (X). 

Also  in  (5)  the  area  is  positive,  in  (6)  negative.  Q.E.D. 

An  easy  way  to  apply  (X)  is  given  by  the  following 

Rule  for  finding  the  area  of  a  triangle. 

First  step.    Write   down   the   vertices   in   two   columns,      «i    y\ 
abscissas   in  one,   ordinates  in   the  other,   repeating   the     x2    y 
coordinates  of  the  first  vertex.  £1    y\ 

Second  step.  Multiply  each  abscissa  by  the  ordinate  of  the  next 
row,  and  add  results.  This  gives  x^yz  -f  X2y8  4-  x^. 

Third  step.  Multiply  each  ordinate  by  the  abscissa  of  the  next 
row,  and  add  results.  This  gives  y^x^  +  y»x8  -f-  yzx^. 

Fourth  step.  Subtract  the  result  of  the  third  step  from  that  of 
the  second  step,  and  divide  by  2.  This  gives  the  required  area, 
namely,  formula  (X). 


38 


ANALYTIC  GEOMETRY 


It  is  easy  to  show  in  the  same  manner  that  the  rule  applies  to 
any  polygon,  if  the  following  caution  be  observed  in  the  first  step : 

Write  down  the  coordinates  of  the  vertices  in  an  order  agreeing 
with  that  established  by  passing  continuously  around  the  perimeter, 
and  repeat  the  coordinates  of  the  first  vertex. 

Ex.  2.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (1,  6), 
(-3,  -4),  (2,  -2),  (-1,3). 

Solution.    Plotting,  we  have  the  figure  from  which 
we  choose  the  order  of  the  vertices  as  indi-         ^ 
cated  by  the  arrows.    Following  the  rule  :     _  i         3 

First  step.    Write  down  the  vertices  in     —  3     —  4 
order. 

Second  step.    Multiply  each   abscissa 
by  the  ordinate  of  the  next  row,  and  add.    This  gives 

1  x  3  +  (- 1  x  -  4)  +  (-  3  x  -  2)  +  2  x  6  =  25. 

Third  step.    Multiply  each  ordinate  by  the  abscissa 
of  the  next  row  and  add.     This  gives 

6  x  -1  +  3  x  -  3  +  (-  4  x  2)  +  (-  2  x  1)  =  -  25. 

Fourth  step.    Subtract  the  result  of  the  third  step 
from  the  result  of  the  second  step,  and  divide  by  2. 

.•.  Area  =  —        —  =  25  unit  squares.     -4ns. 
2 

The  result  has  the  positive  sign,  since  the  area  is  on  the  left. 


PROBLEMS 

1.  Find  the  area  of  the  triangle  whose  vertices  are  (2,  3),  (1,  5),  (—  1,  —  2). 

Ans.  i£. 

2.  Find  the  area  of  the  triangle  whsSe" vertices  are  (2,  3),  (4,  -5),  (-3,  -6). 

Ans.  29. 

3.  Find  the  area  of  the  triangle  whose  vertices  are  (8,  3),  (—2,  3),  (4,  —  5). 

Ans.  40. 

4.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  0),  (-  a,  0),  (0,  6). 

Ans.  ab. 

6.  Find  the  area  of  the  triangle  whose  vertices  are  (0,  0),  (xi,  y\),  (x2,  yz). 

Ans. 


CARTESIAN  COORDINATES  39 

6.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  1),  (0,  6),  (c,  1). 


7.  Find  the  area  of  the  triangle  whose  vertices  are  (a,  6),  (6,  a),  (c,  —  c). 

Ans. 


8.  Find  the  area  of  the  triangle  whose  vertices  are  (3,  0)  ,  (0,  3  V§),  (6,  3  V3). 

Ans.  9V3. 

9.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(2,  3),  (5,  4),  (—4,  1)  is  zero,  and  hence  that  these  points  all  lie  on  the  same 
straight  line. 

10.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(a,  b  -f  c),  (6,  c  +  a),  (c,  a  +  6)  is  zero,  and  hence  that  these  points  all  lie  on 
the  same  straight  line. 

11.  Prove  that  the  area  of  the  triangle  whose  vertices  are  the  points 
(a,  c  +  a),  (—  c,  0),  (—  a,  c  —  a)  is  zero,  and  hence  that  these  points  all  lie 
on  the  same  straight  line. 

12.  Find  the  area  of  the  quadrilateral  whose  vertices  are   (—2,  3), 
(-3,   -4),  (5,   -1),  (2,  2).  Ans.  31. 

13.  Find  the  area  of  the  pentagon  whose  vertices  are  (1,  2),  (3,    —  1), 
(6,  -  2),  (2,  5),  (4,  4).  Ans.   18. 

14.  Find  the  area  of  the  parallelogram  whose  vertices  are  (10,  5),  (—2,  5), 
(_  5,  _  3),  (7,  _  3).  Ans.  96. 

15.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (0,  0),  (5,  0), 
(9,  11),  (0,  3).  Ans.  41. 

16.  Find  the  area  of  the  quadrilateral  whose  vertices  are  (7,  0),  (11,  9), 
(0,  5),  (0,  0).  '    Ans.  59. 

17.  Show  that  the  area  of  the  triangle  whose  vertices  are  (4,  6),  (2,.  —  4), 
(—  4,  2)  is  four  times  the  area  of  the  triangle  formed  by  joining  the  middle 
points  of  the  sides. 

18.  Show  that  the  lines  drawn  from  the  vertices  (3,  —  8),  (—  4,  6),  (7,  0) 
to  the  medial  point  of  the  triangle  divide  it  into  three  triangles  of  equal  area. 

19.  Given  the  quadrilateral  whose  vertices  are  (0,  0),  (6,  8),  (10,  -  2), 
(4,  —  4)  ;   show  that  the  area  of  the  quadrilateral  formed  by  joining  the 
middle  points  of  its  adjacent  sides  is  equal  to  one  half  the  area  of  the  given 
quadrilateral. 


40 


ANALYTIC   GEOMETRY 


23.  Second  theorem  of  projection. 

Lemma  I.    If  Ml}  M2,  Mz  are  any  three  points  on  a  directed  line, 
then  in  all  cases 


Proof.    Let  0  be  the  origin. 

By  (1),  p.  17,  M^Mt  =  OM2  -  OMlf 

MZM3  =  OM8  - 

Adding,          M^MZ  -f  MZM3  ==  OMZ  — 
But  by  (1),  p.  17,         J^A/g  =  OM3  — 
.'.  M^MZ  =  MJI*  -f 


Q.B.D. 


This  result  is  easily  extended  to  prove 

Lemma  II.    If  MI,  Mz,  M8,  -  -  •,  MH_19  Mn  are  any  n  points  on  a 
directed  line,  then  in  all  cases 


the  lengths  in  the  right-hand  member  being  so  written  that  the 
second  point  of  each  length  is  the  first  point  of  the  next. 

The  line  joining  the  first  and  last  points  of  a  broken  line  is 
called  the  closing  line. 


O  M 


(1) 


\2) 


Thus  in  Fig,  1  the  closing  line  is  PiP8 ;  in  Fig.  2  the  closing 
line  is  PiP6- 


CARTESIAN  COORDINATES 


41 


Theorem  XI.  Second  theorem  of  projection.  If  each  segment  of  a 
broken  line  be  given  the  direction  determined  in  passing  continuously 
from  one  extremity  to  the  other,  then  the  algebraic  sum  of  the  pro- 
jections of  the  segments  upon  any  directed  line  equals  the  projection 
of  the  closing  line. 

Proof.  The  proof  results  immediately  from  the  Lemmas.  For 
in  Fig.  1 

M^MI  =  projection  of  PiP2  > 
MZM3  =  projection  of  P2P3 ; 
M^MS  =  projection  of  closing  line  PiP8. 

But  by  Lemma  I 


Q.E.D. 


and  the  theorem  follows. 
Similarly  in  Fig.  2. 

Corollary.  If  the  sides  of  a  closed  polygon  be  given  the  direction 
established  by  passing  continuously  around  the  perimeter,  the  sum 
of  the  projections  of  the  sides  upon  any  directed  line  is  zero. 

For  the  closing  line  is  now  zero. 

Ex.  1.    Find  the  projection  of  the  line  joining  the  origin  and  (5,  3)  upon 

a  line  passing  through  (  —  5,  0)  whose 

inclination  is  —  • 
4 

Solution.    In  the  figure,  applying  the 
second  theorem  of  projection, 

proj.  of  OP  on  A  B 

=  proj.  of  OM  +  proj.  of  MP 

=  OMcos  -  +  MP  cos  - 
4  4 

(by  first  theorem  of  projection,  p.  23) 

Ana. 


r/ 

\ 

H 

/ 

X 

X 

M, 

¥ 

\ 

/ 

\ 

\ 

/ 

\ 

\ 

^ 

\ 

x 

s 

/ 

r 

\ 

^ 

\ 

A 

s 

^ 

/ 

(-51  0) 

O 

M 

The  essential  point  in  the  solution  of  problems  like  Ex.  1  is  the  replacing 
of  the  given  line,  by  means  of  Theorem  XI,  by  a  broken  line  with  two  seg- 
ments which  are  parallel  to  the  axes. 


42 


ANALYTIC  GEOMETRY 


Ex.  2.    Find  the  perpendicular  distance  from  the  line  passing  through 

o  _, 

(4,  0),  whose  inclination  is  — ,  to  the 
point  (10,  2). 

Solution.    In  the  figure  draw  OC 
perpendicular  to  the  given  line  AB. 

=  — ,  or  120°. 


=  30°,  Z  ^OF  =  60°. 

Required  the  perpendicular  dis- 
tance RP. 

Project  the  broken  line  OMP  upon 
0(7.  Then,  by  the  second  theorem  of 

projection, 


{\ 

\ 

\ 

£ 

\ 

/ 

^ 

N^ 

f. 

j? 

\ 

, 

,,' 

\ 

no 

2) 

7T 

2 

\ 

_^  —  - 

-^r" 

-—  ' 

-', 

f 

^ 

_-^-~- 

_-  —  • 

5 

sf 

A 

V' 

TT, 

.!/ 

lY 

tf> 

\ 

! 

proj.  of  OP  =  proj.  of  OM  +  proj.  of  MP 

=  OMcos  Z  XOS  +  MP  cos  Z  SOY 


P)  =  1  +  5  VS. 

.But  in  the  figure 

proj.  of  OP  =  OS  +  ST 

=  OA  cos  JTOS  + 

(2)  =  4  •  !•  V3  +  RP. 

From  (1)  and  (2), 

RP  +  2  V§  =  1  +  5  V|. 

EP  =  1  +  3  Vs.     Ans. 


PROBLEMS 

1.  Four  points  lie  on  the  axis  of  abscissas  at  distances  of  1,  3,  6,  and  10 
respectively  from  the  origin.     Find  P\P±  by  Lemma  II. 

2.  A  broken  line  joins  continuously  the  points  (-  1,  4),  (3,  6),  (6,  -  2), 
(8,  1),  (1,  —  1).    Show  that  the  second  theorem  of  projection  holds  when  the 
segments  are  projected  on  the  X-axis. 

3.  Show  by  means  of  a  figure  that  the  projection  of  the  broken  line  join- 
ing the  points  (1,  2),  <5,  4),  (-  1,  -  4),  (3,  -  1),  and  (1,  2)  upon  any  line  is 
zero. 

4.  Find  the  projection  of  the  line  joining  the  points  (2,  1)  and  (5,  3)  upon 


a  line  passing  through  the  point  (—1,  1)  whose  inclination  is  — . 


Ans. 


3  V3 


CARTESIAN  COORDINATES  43 

5.  What  is  the  projection  of  the  line  joining  these  same  points  upon  any 
line  whose  inclination  is  -  ?     Why  ? 

6.  Find  the  projection  of  the  line  joining  the  points  (—  1,  3)  and  (2,  4) 
upon  any  line  whose  inclination  is  £  it.  Ans.   —  V2. 

7.  Find  the  projection  of  the  broken  line  joining  the  points  (—1,  4), 

(3,  6),  and  (5,  0)  upon  a  line  whose  inclination  is  —  •     Verify  your  result  by 
finding  the  projection  of  the  closing  line.  Ans.   Vi. 

8.  Find  the  projection  of  the  broken  line  joining  (0,  0),  (4,  2),  and  (6,  —  3) 

2  TIT  A      Q  *\XQ 

upon  a  line  whose  inclination  is  —^-  •  Ans     ~          — _. 

8  2 

~"  9.  Show  that  the  projection  of  the  sides  of  the  triangle  (2,  1),  (—1,  5), 
(—3,  1)  upon  a  line  whose  inclination  is  —  is  zero. 

10.  Find  the  perpendicular  distance  from  the  point  (6,  3)  to  a  line  passing 

through  the  point  (—4,  0)  with  an  inclination  of  —  •  Ans.   —=• 

4  V2 

11.  Find  the  perpendicular  distance  from  the  point  (—5,  —  1)  to  a  line 
passing  through  the  point  (6,  0)  and  having  an  inclination  of  £  it. 

Ans.   6V2. 

12.  A  line  of  inclination  —  passes  through  the  point  (5,  0).    Find  the  per- 
pendicular distance  to  the  parallel  line  passing  through  the  point  (0,  2). 

5  +  2  V3 
Ans. • 


CHAPTER  III 
THE  CURVE  AND  THE  EQUATION 

24.  Locus  of  a  point  satisfying  a  given  condition.   The  curve* 
(or  group  of  curves)  passing  through  all  points  which  satisfy  a 
given  condition,  and  through  no  other  points,  is  called  the  locus 
of  the  point  satisfying  that  condition. 

For  example,  in  Plane  Geometry,  the  following  results  are 
proved  : 

The  perpendicular  bisector  of  the  line  joining  two  fixed  points 
is  the  locus  of  all  points  equidistant  from  these  points. 

The  bisectors  of  the  adjacent  angles  formed  by  two  lines  is 
the  locus  of  all  points  equidistant  from  these  lines. 

To  solve  any  locus  problem  involves  two  things : 

1.  To  draw  the  locus  by  constructing  a  sufficient  number  of 
points  satisfying  the  given  condition  and  therefore  lying  on  the 
locus. 

2.  To  discuss  the  nature  of  the  locus,  that  is,  to  determine 
properties  of  the  curve,  f 

Analytic  Geometry  is  peculiarly  adapted  to  the  solution  of 
both  parts  of  a  locus  problem. 

25.  Equation  of  the  locus  of  a  point  satisfying  a  given 
condition.    Let  us  take  up  the  locus  problem,  making  use  of  coor- 
dinates.    If  any  point  P  satisfying  the  given  condition  and  there- 
fore lying  on  the  locus  be  given  the  coordinates  (x,  y),  then  the 
given  condition  will  lead  to  an  equation  involving  the  variables 
x  and  y.     The  following  example  illustrates  this  fact,  which  is 
of  fundamental  importance. 

*  The  word  "  curve  "  will  hereafter  signify  any  continuous  line,  straight  or  curved. 

t  As  the  only  loci  considered  in  Elementary  Geometry  are  straight  lines  and  circles, 
the  complete  loci  may  be  constructed  by  ruler  and  compasses,  and  the  second  part  i3 
relatively  unimportant. 

44 


THE  CURVE  AND  THE  EQUATION 


45 


Ex.  1.    Find  the  equation  in  x  and  y  if  the  point  whose  locus  is  required 
,11  be  equidistant  from  A  (—  2,  0)  and  .B(—  3,  8). 

Solution.   Let  P  (x,  y)  be  any  point  on  the  locus.  Then  by  the  given  condition 
(1)  PA  =  PB. 

But,  by  formula  IV,  p.  24, 


(2) 


f  ^        PJ.  =  V(x  +  2)a  +  (y-Q)8, 

P£  =  V(x  +  3)2  +  (y  -  8)2. 

Substituting  in  (1), 

V(x  +  2)2  +  (y  -  0)2 


=  V(x  +  3)*  +  (y  -  8)2. 
Squaring  and  reducing, 
(3)  2  x  -  16  y  +  69  =  0. 

In  the  equation  (3),  x  and  y  are  variables  representing  the  coordinates  of 
any  point  on  the  locus,  that  is,  of  any  point  on  the  perpendicular  bisector  of 
the  line  AB.  This  equation  has  two  important  and  characteristic  properties : 

1.  The  coordinates  of  any  point  on  the  locus  may  be  substituted  for  x 
and  y  in  the  equation  (3),  and  the  result  will  be  true. 

For  let  PI  (xi,  7/1)  be  any  point  on  the  locus.  Then  P\A  =  Pi-B,  by  defi- 
nition. Hence,  by  formula  IV,  p.  24, 

(4)  V(zi  +  2)2  +  yia  =  V(zi  +  3)2  +  O/i  -  8)2, 
or,  squaring  and  reducing, 

(5)  2x1-162/1  +  69  =  0. 
Therefore  Xi  and  yi  satisfy  (3). 

2.  Conversely,  every  point  whose  coordinates  satisfy  (3)  will  lie  upon  the 
locus. 

For  if  PI(XI,  2/1)  is  a  point  whose  coordinates  satisfy  (3),  then  (5)  is  true, 
and  hence  also  (4)  holds.  Q.E.D. 

In  particular,  the  coordinates  of  the  middle  point  C  of  A  and  B,  namely, 
z  =  -2i,  y  =  4  (Corollary,  p.  32),  satisfy  (3),  since  2 (-2£) -16  x  4  +  69  =  0. 

This  example  illustrates  the  following  correspondence  between 
Pure  and  Analytic  Geometry  as  regards  the  locus  problem : 

Locus  problem 

Pure  Geometry  Analytic  Geometry 

The  geometrical  condition  (satis-      An  equation  in  the  variables  x 


fied  by  every  point  on  the 
locus). 


and  y  representing  coordinates 
(satisfied  by  the  coordinates 
of  every  point  on  the  locus). 


46 


ANALYTIC  GEOMETRY 


This  discussion  leads  to  the  fundamental  definition  : 

The  equation  of  the  locus  of  a  point  satisfying  a  given  condition 
is  an  equation  in  the  variables  x  and  y  representing  coordinates 
such  that  (1)  the  coordinates  of  every  point  on  the  locus  will 
satisfy  the  equation;  and  (2)  conversely,  every  point  whose 
coordinates  satisfy  the  equation  will  lie  upon  the  locus. 

This  definition  shows  that  the  equation  of  the  locus  must  be 
tested  in  two  ways  after  derivation,  as  illustrated  in  the  example 
of  this  section  and  in  those  following. 

From  the  above  definition  follows  at  once  the 

Corollary.  A  point  lies  upon  a  curve  when  and  only  when  its 
coordinates  satisfy  the  equation  of  the  curve. 

26.  First  fundamental  problem.  To  find  the  equation  of  a 
curve  which  is  defined  as  the  locus  of  a  point  satisfying  a  given 
condition. 

The  following  rule  will  suffice  for  the  solution  of  this  problem 
in  many  cases : 

Rule.  First  step.  Assume  that  P  (x,  y)  is  any  point  satisfying 
the  given  condition  and  is  therefore  on  the  curve. 

Second  step.     Write  down  the  given  condition. 

Third  step.  Express  the  given  condition  in  coordinates  and 
simplify  the  result.  The  final  equation,  containing  x,  y,  and  the 
given  constants  of  the  problem,  will  be  the  required  equation. 

Ex.  1.  Find  the  equation  of  the  straight  line  passing  through  P!  (4,  —  1) 
and  having  an  inclination  of 

Solution.  First  step.  Assume  P(z,  y)  any  point 
on  the  line. 

Second  step.  The  given  condition,  since  the  incli- 
nation a.  is  —  i  may  be  written 


4-  1 


(1)  Slope  of  PiP  =  tan  a  =  —  ll 

Third  step.    From  (V),  p.  28, 


NS 

\ 

1 

sQ 

w 

-)- 

\ 

3 

O 

\ 

<%• 

i) 

PJ 

\ 

(2) 


Slope  of  PiP  =  tan  a  = 


3/1  -2/2 
Xi  —x* 


x  —  4 
[By  substituting  (x,  y)  for  (xlf  yx),  and  (4,  —  Ti  for  (x2, 


THE  CURVE  AND  THE  EQUATION 


from  (1), 


47 


x  +  y  —  3  =  0.     Ans. 


To  prove  that  (3)  is  the  required  equation : 

1.  The  coordinates  (x1?  ?/i)  of  any  point  on  the  line  will  satisfy  (3),  for 
the  line  joins  (xi,  2/1)  and  (4,  -  1),  and  its  slope  is  —  1 ;  hence,  by  (V),  p.  28, 
substituting  (4,  —  1)  for  (x2,  2/2), 


or 


—  3  =  0, 


and  therefore  Xi  and  2/1  satisfy  the  equation  (3). 

2.  Conversely,  any  point  whose  coordinates  satisfy  (3)  is  a  point  on  the 
straight  line.     For  if  (xi,  2/1)  is  any  such  point,  that  is,  if  Xi  +  2/1  —  3  =  0,  then 

also  —  1  =  —  -  is  true,  and  (xi,  2/1)  is  a  point  on  the  line  passing  through 


(4,  —  1)  and  having  an  inclination  equal  to  —r-m 


Q.E.D. 


Ex.  2.    Find  the  equation  of  a  straight  line  parallel  to  the  axis  of  y  and 
at  a  distance  of  6  units  to  the  right. 

Solution.  First  step.  Assume  that  P(x,  y)  is 
any  point  on  the  line,  and  draw  NP  perpendicular 
to  OF. 

Second  step.  The  given  condition  may  be 
written 

(4)  NP  =  6. 

Third  step.    Since  NP  =  OM  =  x,  (4)  becomes 

(5)  x  =  G.     Ans. 


tt 

(x 

y) 

0 

M 

X 

The  equation  (5)  is  the  required  equation  : 

1.  The  coordinates  of  every  point  satisfying  the  given  condition  may  be 
substituted  in  (5).     For  if  PI  (xi,  yi)  is  any  such  point,  then  by  the  given 
condition  x\  =  6,  that  is,  (xi,  j/i)  satisfies  (5). 

2.  Conversely,  if  the  coordinates  (xi,  yi)  satisfy  (5),  then  Xi  =  6,  and 
•Pi(*i>  2/i)  is  at  a  distance  of  six  units  to  the  right  of  YY'.  Q.E.D. 

The  method  above  illustrated  of  proving  that  the  derived  equation  has  the 
two  characteristic  properties  of  the  equation  of  the  locus  should  be  carefully 
studied  and  applied  to  each  of  the  following  examples. 


48 


ANALYTIC  GEOMETRY 


PC  =  V(x  +  I)2  +  (y  -  2)2. 
Substituting  in  (6), 


V(x  +  I)2  +  (y  -  2)2  =  4. 


Squaring  and  reducing, 


1.2) 


Ex.  3.    Find  the  equation  of  the  locus  of  a  point  whose  distance  from 
(—1,  2)  is  always  equal  to  4. 

Solution.  First  step.  Assume  that  P(x,y) 
is  any  point  on  the  locus. 

Second  step.    Denoting  (—  1,  2)  by  C, 
the  given  condition  is 
(6)  PC  =  4. 

Third  step.    By  formula  (IV),  p.  24, 


(7) 


2, 


2x_4y-  11  =  0. 


!/) 


X 


This  is  the  required  equation,  namely,  the  equation  of  the  circle  whose 
center  is  (—1,  2)  and  radius  equals  4.  The  method  of  proof  is  the  same 
as  that  of  the  preceding  examples. 


PROBLEMS 

1.  Find  the  equation  of  a  line  parallel  to  OY  and 

(a)  at  a  distance  of  4  units  to  the  right. 

(b)  at  a  distance  of  7  units  to  the  left. 

(c)  at  a  distance  of  2  units  to  the  right  of  (3,  2). 

(d)  at  a  distance  of  5  units  to  the  left  of  (2,  —  2). 

2.  What  is  the  equation  of  a  line  parallel  to  OF  and  a  —  6  units  from  it  ? 
How  does  this  line  lie  relative  toOIrifa>&>0?   if  0  >  6  >  a  ? 

3.  Find  the  equation  of  a  line  parallel  to  OX  and 

(a)  at  a  distance  of  3  units  above  OX. 

(b)  at  a  distance  of  6  units  below  OX. 

(c)  at  a  distance  of  7  units  above  (—2,  —  3). 

(d)  at  a  distance  of  5  units  below  (4,  —  2). 

4.  What  is  the  equation  of  XX'  ?  of  YY'  ? 

5.  Find  the  equation  of  a  line  parallel  to  the  line  x  =  4  and  3  units  to  the 
right  of  it.     Eight  units  to  the  left  of  it.     • 

6.  Find  the  equation  of  a  line  parallel  to  the  line  y  =  -  2  and  4  units 
below  it.     Five  units  above  it. 

7.  How  does  the  line  y  =  a  -  b  lie  if  a  >'b  >  0?   if  6  >  a  >  0? 

8.  What  is  the  equation  of  the  axis  of  x  ?  of  the  axis  of  y  ? 


THE  CURVE  AND  THE  EQUATION         49 

9:  What  is  the  equation  of  the  locus  of  a  point  which  moves  always  at 
a  distance  of,  2  units  from  the  axis  of  x  ?  from  the  axis  of  y  ?  from  the  line 
x  =  —  5  ?  from  the  line  y  =  4  ? 

10.  What  is  the  equation  of  the  locus  of  a  point  which  moves  so  as  to 
be  equidistant  from  the  lines  x  =  5  and  x  =  9  ?  equidistant  from  y  =  3  and 
y=-7? 

11.  What  are  the  equations  of  the  sides  of  the  rectangle  whose  vertices 
are  (5,  2),  (5,  5),  (- 2,  2),  (- 2,  5)  ? 

In  problems  12  and  13,  PI  is  a  given  point  on  the  required  line,  m  is  the 
slope  of  the  line,  and  a  its  inclination. 

12.  What  is  the  equation  of  a  line  if 

(a)  PI  is  (0,  3)  and  m  =  —  3 ?  Ans.  3x  +  y  —  3  =  0. 

(b)  PI  is  (—  4,  —  2)  and  m  —  \  ?  Ans.  x-3y  —  2  =  0. 

(c)  P!  is  (-  2,  3)  andm  =  — -?  Ans.  V<ix  -2y  +  6  +  2\/2  =  0. 

Vs 

(d)  P!  is  (0,  5)  and  m  = ?  Ans.  V3x  -  2y  +  10  =  0. 

(e)  P!  is  (0,  0)  and  m  =  -  |  ?  Ans.  2x  +  3y  =  0. 

(f )  PI  is  (a,  6)  and  m  =  0  ?  Ans.  y  =  b. 

(g)  PI  is  (—  a,  6)  and  m  =  oo  ?  Ans.  x  =  —  a. 

13.  What  is  the  equation  of  a  line  if 

J(  (a)  PI  is  (2,  3)  and  a  =  45°  ?  Ans.  x-y  +  1  =  0. 

(b)  P!  is  (-1,  2)  and  a  =  45°?  Ans.  x  -  y  +  3  =  0. 

(c)  PI  is  ( -  a,  -  b)  and  a  =  45°  ?  4ns.  x  -  y  =  b  -  a. 

(d)  P!  is  (5,  2)  and  a  =  60°?  4ns.   V3x  -  y  +  2  -  5  VJJ  =  0. 

(e)  P!  is  (0,  -  7)  and  a.  =  60°?  Ans.  V§x  -  y  -  7  =  0. 

(f)  Pi  is  (-  4,  5)  and  a  =  0° ?  .Ans.  y  =  5. 

(g)  PI  is  (2,  -  3)  and  a  =  90°  ?  Ans.  x  =  2. 

(h)  P!  is  (3,  -  3  V§)  and  or  =  120°?  4ns.   V3x  +  y  =  0. 

(i)  PI  is  (0,  3)  and  a  =  150°?  Ans.   V3x  +  3y  -  9  =  0. 

(j)  PI  is  (a,  6)  and  a  =  135°  ?  4ns.  x  +  y  =  a  4-  6. 
/V\-      K 

14.  Are  the  points  (3,  9),  (4,  6),  (5,  5)  on  the  line  3z  +  2y  =  25? 

15.  Find  the  equation  of  the  circle  with 

(a)  center  at  (3,  2)  and  radius  =  4.  4ns.  x2  +  ?/2-6z-4y-3  =  0. 

(b)  center  at  (12,  -  5)  and  r  =  13.  4ns.  x2  +  y2  —  24  x  +  10  y  =  0. 

(c)  center  at  (0,  0)  and  radius  =  r.  Ans.  x2  +  y2  =  r2. 

(d)  center  at  (0,  0)  and  r  =  5.  4ns.  x2  +  y2  =  25. 

(e)  center  at  (3  a,  4  a)  and  r  =  5  a.  Ans.  x2  +  y2  -  2  a  (3  x  +  4  y)  =  0. 

(f)  center  at  (6  +  c,  6  —  c)  and  r  =  c, 

4ns.  x2  +  y2  -  2 (fr  +  c)x  -  2 (6  -  c)y  +  2 62  +  c2  =  (X 


50  ANALYTIC  GEOMETRY 

16.  Find  the  equation  of  a  circle  whose-  center  is  (5,  —  4)  and  whose 
circumference  passes  through  the  point  (—2,  3). 

17.  Find  the  equation  of  a  circle  haying  the  line  joining  (3,  —  5)  and 
(—  2,  2)  as  a  diameter. 

18.  Find  the  equation  of  a  circle  touching  each  axis  at  a  distance  6  units 
from  the  origin. 

19.  Find  the  equation  of  a  circle  whose  center  is  the  middle  point  of  the 
line  joining  (—6,  8)  to  the  origin  'and  whose  circumference  passes  through 
the  point  (2,  3). 

20.  A  point  moves  so  that  its  distances  from  the  two  fixed  points  (2,  —  3) 
and  (—1,  4)  are  equal.     Find  the  equation  of  the  locus  and  plot. 

Ans.   3x-7y  +  2  =  0. 

21.  Find  the  equation  of  the  perpendicular  bisector  of  the  line  joining 

(a)  (2,  1),  (-  3,  -  3).  Ans.  10  x  +  8  y  +  13  =  0. 

(b)  (3,  1),  (2,  4).  Ans.  a-3y  +  6  =  0. 

(c)  (-  1,  -  1),  (3,  7).  Ana.  x+2y-7  =  0. 

(d)  (0,  4),  (3,  0).  Ans.  Qx  -  Sy  +  7  =  0. 

(e)  (xi,  2/i),  (x2,  2/2). 

Ans.  2  (xi  -  z2)x  +  2  (y^  -y2)y  +  x22  -  Xi2  +  ?/22  -  j/i2  =  0. 


22.  Show  that  in  problem  21  the  coordinates  of  the  middle  point  of 
the  line  joining  the  given  points  satisfy  the  equation  of  the  perpendicular 
bisector. 

23.  Find  the  equations  of  the  perpendicular  bisectors  of  the  sides  of  the 
triangle  (4,  8),  (10,  0),  (6,  2).     Show  that  they  meet  in  the  point  (11,  7). 

24.  Express  by  an  equation  that  the  point  (ft,  k)  is  equidistant  from 
(—  1,  1)  and  (1,  2)  ;  also  from  (1,  2)  and  (1,  -  2).     Then  show  that  the  point 
(f,  0)  is  equidistant  from  (-  1,  1),  (1,  2),  (1,  -  2). 

27.  General  equations  of  the  straight  line  and  circle.     The 

methods  illustrated  in  the  preceding  section  enable  us  to  state 
the  following  results  : 

1.  A  straight  line  parallel  to  the  axis  of  y  has  an  equation  of 
the  form  x  =  constant. 

2.  A  straight  line  parallel  to  the  axis  of  x  has  an  equation  of 
the  form  y  =  constant. 


THE  CURVE  AND  THE  EQUATION         51 


Theorem  I.    The  equation ,  of  the  straight  line  passing  through  a 
point  B  (0,  b)  on  the  axis  of  y  and  having  its  slope  equal  to  in  is 

(I)  \^jT  =  mv  +  &.~^ 

Proof.    First  step.    Assume  that  P  (x,  y)  is  any  point  on  the  line. 
Second  step.    The  given  condition  may  be  written 

Slope  of  PB  —  m. 
Third  step.    Since  by  Theorem  V,  p.  28, 

Slope  of  PB  =  y^^, 
x  —  0 

[Substituting  (x,  y)  for  (xi,  yi)  and  (0,  6)  for  (z2,  2/2)] 

y-l 

then  =  m,  or  y  =  mx  +  £.  Q.E.D. 

x 

Theorem  II.    The  equation  of  the  circle -whose  center  is  a  given 
point  (a,  ft)  and  whose  radius  equals  r  is 

(II)  a?2  +  7/2  -  2  an  -  2  $y  +  a2  +  £2  -  rz  =  O. 

Proof.    First  step.    Assume  that  P(x,  ?/)  is  any  point  on  the 
locus. 

Second  step.    If  the  center  (a,  ft)  be 'denoted  by  C,  the  given 

condition  is 

PC  =  r. 

Third  step.    By  (IV),  p.  24, 


PC  =  V(a?  -  a)2  +  (y  -  /?)2. 
.'.  V(re  —  a)2  +  (y  —  /2)2  =  r. 
Squaring  and  transposing,  we  have  (II).  Q.E.D. 

Corollary.    The  equation  of  the  circle  whose  center  is  the  origin 
(0,  0)  and  whose  radius  is  r  is 


The  following  facts  sKpj;ld  be  observed : 

Any  straight  line  is  defined  by  an  equation,  of  the  first  degree 
in  the  variables  x  and  y. 

Any  circle  is  defined  by  an  equation  of  the  second  degree  in 
the  variables  x  and  ?/,  in  which  the  terms  of  the  second  degree 
consist  of  the  sum  of  the  squares  of  x  and  y. 


52  ANALYTIC  GEOMETRY 

28.  Locus  of  an  equation.  The  preceding  sections  have  illus- 
trated the  fact  that  a  locus  problem  in  Analytic  Geometry  leads 
at  once  to  an  equation  in  the  variables  x  and  y.  This  equation 
having  been  found  or  being  given,  the  complete  solution  of  the 
locus  problem  requires  two  things,  as  already  noted  in  the  first 
section  (p.  44)  of  this  chapter,  namely, 

1.  To  draw  the  locus  by  plotting  a  sufficient  number  of  points 
whose  coordinates  satisfy  the  given  equation,  and  through  which 
the  locus  therefore  passes. 

2.  To  discuss  the  nature  of  the  locus,  that  is,  to  determine 
properties  of  the  curve. 

These  two  problems  are  respectively  called : 

1.  Plotting   the   locus  of    an    equation   (second   fundamental 
problem). 

2.  Discussing  an  equation  (third  fundamental  problem). 

For  the  present,  then,  we  concentrate  our  attention  upon  some 
given  equation  in  the  variables  x  and  y  (one  or  both)  and  start 
out  with  the  definition  : 

The  locus  of  an  equation  in  two  variables  representing  coordinates 
is  the  curve  or  group  of  curves  passing  through  all  points  whose 
coordinates  satisfy  that  equation,*  and  through  such  points  only. 

From  this  definition  the  truth  of  the  following  theorem  is  at 
once  apparent : 

Theorem  III.  If  the  form  of  the  given  equation  be  changed  in  any 
way  (for  example,  by  transposition,  by  'multiplication  by  a  constant, 
etc.),  the  locus  is  entirely  unaffected. 

*  An  equation  in  the  variables  x  and  y  is  not  necessarily  satisfied  by  the  coordinates  of 
any  points.  For  coordinates  are  real  numbers,  and  the  form  of  the  equation  may  be  such 
that  it  is  satisfied  by  no  real  values  of  x  and  y.  For  example,  the  equation 

#2  +  7/2+1=0 

is  of  this  sort,  since,  when  x  and  y  are  real  numbers,  a?2  and  y2  are  necessarily  positive 
(or  zero),  and  consequently  x*  +  yz+  1  is  always  a  positive  number  greater  than  or  equal 
to  1,  and  therefore  not  equal  to  zero.  Such  an  equation  therefore  has  no  locus.  The 
expression  "the  locus  of  the  equation  is  imaginary"  is  also  used. 

An  equation  may  be  satisfied  by  the  coordinates  of  a  finite  number  of  points  only. 
For  example,  a;2  +  ?/3=0  is  satisfied  by  x=0,  y  =  0,  but  by  no  other  real  values.    In  this 
case  the  group  of  points,  one  or  more,  whose  coordinates  satisfy  the  equation,  is  called^ 
the  locus  of  the  equation. 


53 


THE  CURVE  AND  THE  EQUATION 

We  now  take  up  in  order  the  solution  of  the  second  and  third 
fundamental  problems. 

29.  Second  fundamental  problem. 

Rule  to  plot  the  locus  of  a  given  equation. 

First  step.  Solve  the  given  equation  for  one  of  the  variables  in 
terms  of  the  other.* 

Second  step.  By  this  formula  compute  the  values  of  the  vari- 
able for  which  the  equation  has  been  solved  by  assuming  real 
values  for  the  other  variable. 

Third  step.  Plot  the  points  corresponding  to  the  values  so 
determined.^ 

Fourth  step.  If  the  points  are  numerous  enough  to  suggest  the 
general  shape  of  the  locus,  draw  a  smooth  curve  through  the  points. 

Since  there  is  no  limit  to  the  number  of  points  which  may  be 
computed  in  this  way,  it  is  evident  that  the  locus  may  be  drawn 
as  accurately  as  may  be  desired  by  simply  plotting  a  sufficiently 
large  number  of  points. 

Several  examples  will  now  be  worked  out  and  the  arrangement 
of  the  work  should  be  carefully  noted. 

Ex.  1.    Draw  the  locus  of  the  equation 

2x-3y  +  6  =  0. 
Solution.    First  step.    Solving  for  y, 


30) 


Second  step.    Assume  values  for  x  and  compute 
arranging  results  in  the  form  : 


Thus,  if 

x  =  1,  y  =  I  •  1  +  2  =  2f , 

x  =  2,  y  =  f-2  +  2  =  8*. 

etc. 

Third  step.    Plot  the  points  found. 
Fourth  step.     Draw    a   smooth    curve 
through  these  points. 

*  The  form  of  the  given  equation  will  often  be  such  that  solving  for  one  variable  is 
simpler  than  solving  for  the  other.     Always  choose  the  simpler  solution. 
t  Remember  that  real  values  only  may  be  used  as  coordinates. 


X 

y 

X 

y 

0 

2 

0 

2 

1 

2  f 

-  1 

H 

2 

3J 

-2 

1 

3 

4 

-3 

0 

4 

4| 

-4 

-* 

etc. 

etc. 

etc. 

etc. 

54 


ANALYTIC  GEOMETRY 


Ex.  2.    Plot  the  locus  of  the  equation 

y  =  z2-2x-3. 

Solution.    First  step.    The  equation  as  given  is  solved  for  y. 
Second  step.    Computing  y  by  assuming  values  of  x,  we  find  the  table  of 
values  below : 


X 

y 

X 

V 

0 

-3 

0 

-3 

1 

-4 

-  1 

0 

2 

-3 

-2 

5 

3 

0 

-3 

12 

4 

5 

-4 

21 

5 

12 

etc. 

etc. 

6 

21 

etc. 

etc. 

Third  step.    Plot  the  points. 

Fourth  step.    Draw  a  smooth  curve  through  these  points.     This  gives  the 
curve  of  the  figure. 

Ex.  3.    Plot  the  locus  of  the  equation 


First  step.    Solving  for  ?/, 


y  =  ±Vl6  -6x-x2. 
Second  step.    Compute  y  by  assuming  values  of  X. 


X 

y 

a? 

y 

0 

±4 

0 

±4 

1 

±3 

-  1 

±4.6 

2 

0 

-2 

±4.9 

3 

imag. 

-3 

±5 

4 

t( 

-4 

±4.9 

5 

K 

-5 

±4.6 

6 

u 

-6 

±4 

7 

(I 

Y 

±3 

-8 

0 

-9 

imag. 

Y' 


\ 


X 


THE  CURVE  AND  THE  EQUATION         65 


For  example,  if  x  =  1,  y  =  ±     lG  -  G  -  1  =  ±  3  ; 


if  x  =  3,  y  =  ±  vl6  -  18  -  9  =  ±  V- 11, 
an  imaginary  number ; 

if  x  =  -  1,  y  =  ±  Vl6  +  G  -  1  =  ±  4.6, 
etc. 

Third  step.    Plot  the  corresponding  points. 

Fourth  step.    Draw  a  smooth  curve  through  these  points. 

PROBLEMS 

1.  Plot  the  locus  of  each  of  the  following  equations. 

(a)  x  +  2  y  =  0.  (p)  x2  +  y2  =  9. 

(b)  x  +  2  y  =  3.  (q)  x2  +  y2  =  25. 

<c)  3x  -  y  +  5  =  0.  (r)  x-2  +  3,2  +  9x  =  0. 

<d)  2/  =  4*2-  (s)  x2  +  772  +  42/  =  0. 

S)  x2  +1^"- 5  =  0.  <U)  "2  +  ^  -  Qy  ~  1G 

(h)  2/-*2  +  x  +  l.  (v>  *y  =  «*~8. 

(i)  x  =  2/2  +  2?y  -  3.  <w)  4x  =  ^  +  8' 

(3)4x^7/3.  (X)?y==_^ — 
(k)  4x  =  2/3  -  1.  l  +  x'2 


(m)  y  =  x3  -  x.  1  +  y2 

(n)  7y  =  x3-x2-5.  ,,  g=      2 

(o)  x2  +  y2  =  4.  1  +  2/2 

2.  Show  that  the  following  equations  have  no  locus  (footnote,  p.  52). 

(a)  x2  +  2/2  +  i  _  o.  (f )  x2  +  2/2  +  2  x  +  2  y  +  3  =  0. 

(b)  2x3  +  32/2  =  -  8.  (g)  4x2  +  2/2  +  8x  +  5  =  0. 

(c)  x2  +  4  =  0.  (h)  y4  +  2  x2  +  4  =  0. 

(d)  x4  +  2/'2  +  8  =  0.  (i)  9x2  +  42/2+18x  +  8i/+15=0. 

(e)  (x  -H  I)2  +  2,2  +  4-0.  (j)  x2  +  xy  +  ?/2  +  3  =  0. 

Hint.   Write  each  equation  in  the  form  of  a  sum  of  squares  and  reason  as  in  the  foot- 
x   note  on  p.  52. 

30.  Principle  of  comparison.   In  Ex.  1,  p.  53,  and  Ex.  3,  p.  54, 

we  can  determine  the  nature  of  the  locus,  that  is,  discuss  the  equa- 
tion,  by  making  use  of  the  formulas  (I)  and  (II),  p.  51.  The 
method  is  important,  and  is  known  as  the  principle  of  comparison. 


56  ANALYTIC  GEOMETRY 

The  nature  of  the  locus  of  a  given  equation  may  be  determined 
by  comparison  with  a  general  known  equation,  if  the  latter  becomes 
identical  with  the  given  equation  by  assigning  particular  values  to 
its  coefficients. 

The  method  of  making  the  comparison  is  explained  in  the 
following 

Rule.  First  step.  Change  the  form*  of  the  given  equation  (if 
necessary)  so  that  one  or  more  of  its  terms  shall  be  identical  with 
one  or  more  terms  of  the  general  equation. 

Second  step.  Equate  coefficients  of  corresponding  terms  in  the 
two  equations,  supplying  any  terms  missing  in  the  given  equation 
with  zero  coefficients. 

Third  step.  Solve  the  equations  found  in  tfye  second  step  for 
the  values^  of  the  coefficients  of  the  general  equation. 

Ex.  1.  Show  that  2x  —  3y+6=0  is  the  equation  of  a  straight  line 
(Fig.,  p.  53). 

Solution.    First  step.    Compare  with  the  general  equation  (I),  p.  61, 

(1)  *  y  =  mx  +  6. 

Put  the  given  equation  in  the  form  of  (1)  by  solving  for  y, 

(2)  y=fx  +  2. 

Second  step.  The  right-hand  members  are  now  identical.  Equating 
coefficients  of  x, 

(3)  m  =  f . 
Equating  constant  terms, 

(4)  6  =  2. 

Third  step.  Equations  (3)  and  (4)  give  the  values  of  the  coefficients  m 
and  6,  and  these  are  possible  values,  since,  p.  27,  the  slope  of  a  line  may 
have  any  real  value  whatever,  and  of  course  the  ordinate  b  of  the  point 
(0,  6)  in  which  a  line  crosses  the  F-axis  may  also  be  any  real  number.  There- 
fore the  equation  2x-3y  +  6  =  0  represents  a  straight  line  passing  through 
(0,  2)  and  having  a  slope  equal  to  f .  Q.E.D. 

*  This  transformation  is  called  "  putting  the  given  equation  in  the  form "  of  the 
general  equation. 

tThe  values  thus  found  may  be  impossible  (for  example,  imaginary)  values.  This 
may  indicate  one  of  two  things, — that  the  given  equation  has  no  locus,  or  that  it  cannot 
be  put  in  the  form  required. 


THE  CURVE  AND  THE  EQUATION        57 

Ex.  2.    Show  that  the  locus  of 

(5)  x2  +  y2  +  6x-16  =  0 
is  a  circle  (Fig.,  p.  54). 

Solution.    First  steg.    Compare  with  the  general  equation  (II),  p.  51, 

(6)  x2  +  2/2/-  2  ax  -  20y  +  a2  +  p  -  r2  =  0. 

The  right-hand  members  of  (5)  and  (6)  agree,  and  also  the  first  two  terms, 
X2  +  y2. 

Second  step.    Equating  coefficients  of  x, 

(7)  -  2  a  =  Q. 
Equating  coefficients  of  y, 

(8)  -2£  =  0. 
Equating  constant  terms, 

(9)  «2  +  02  -  r2  =  -  16. 
Third  step.    From  (7)  and  (8), 

a  =  -  3,  /3  =  0. 

Substituting  these  values  in  (9)  and  solving  for  r,  we  find 

r2  =  25,  or  r  =  5. 

Since  a,  /3,  r  may  be  any  real  numbers  whatever,  the  locus  of  (5)  is  a 
circle  whose  center  is  (—3,  0)  and  whose  radius  equals  5. 

PROBLEMS 

1.  Plot  the  locus  of  each  of  the  following  equations.  Prove  that  the  locus 
is  a  straight  line  in  each  case,  and  find  the  slope  m  and  the  point  of  inter- 
section with  the  axis  of  y,  (0,  6). 

(a)  2  x  -f  y  —  6  =  0.  Ans.  m  =  —  2,  6  =  6. 

(b)  x  -  3 y  +  8  =  0.  Ans.  m  =  |,  b  =  2g. 

(c)  x-f  2y  =  q  -4ns.  m  =  -  i,  6  =  0.        M         T 

(d)  5 x  —  6 y  —  5  =  0.  Ans.  m  =  '£,  6  =  -  $. 

(e)  i  x  -  f  y  -  I  =  0.  .Ana.  m  =  f,  b  =-  ^. 

(f)  ?  -  y-  -  1  =  0.  Ana.  m  =  f ,  6  =  -  6. 
5       o 

(g)  7  x  -  8  y  =  0.  Ans.   m  =  |,  6  =  0. 

(h)  f  x  -  f  y  -  |  =  0.  Ans.  m  =  f,  6  =  -  1-&. 


58  ANALYTIC  GEOMETRY 

2.  Plot  the  locus  of  each  of  the  equations  following,  and  prove  that  the 
locus  is  a  circle,  finding  the  center  (a,  /3)  and  the  radius  r  in  each  case. 

(a)  x2  +  2/2  _  i6  =  o.  Ans.  (a,  0)  =  (0,  0);  r  =  4. 

(b)  x2  +  2/2  -  49  =  0.  Ans.   (a,  ft  =  (0,  0);  r  =  7. 

(c)  x2  +  y2  -  25  =  0.  .4ns.  (a,  £)  =  (0,  0)  ;  r  =  5. 

(d)  x2  +  2/2  +  4x  =  0.  .4ns.   (a,  /3)  =  (-  2,  0);  r  =  2. 

(e)  x2  +  y2  -  8  y  =  0.  4ns.  (a,  0)  =  (0,  4);  r  =  4. 

(f)  x2  +  y2  +  4x  -  8y  =  0.  4ns.   (or,  0)  =  (-  2,  4);  r  =  V20. 

(g)  x2  +  y2  -  6x  +  4y  -  12  =  0.  Ans.  (a,  0)  =  (3,  -  2);  r  =  5. 
(h)  x2  +  y2  -  4x  +  9y  -  |  =  0.  Ans.   (a,  /3)  =  (2,  -  f);  r  =  5. 

(i)  3x2  +  3?/2  -  6x  -  8y  =  0.  4ns.   (a,  /3)  =  (1,|);  r  =  f  . 


The  following  problems  illustrate  cases  in  which  the  locus 
problem  is  completely  solved  by  analytic  methods,  since  the  loci 
may  be  easily  drawn  and  their  nature  determined. 

3.  Find  the  equation  of  the  locus  of  a  point  whose  distances  from  the 
axes  XX'  and  YY'  are  in  a  constant  ratio  equal  to  f  . 

2t  -v  'Su.  Ans.    The  straight  line  2  x  —  3  y  =  0. 

4.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  whose  distances 
from  the  axes  of  coordinates  is  always  equal  to  10. 

-?vo  Ans.   The  straight  line  x  +  y  —  10  =  0. 


5.  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  (3,  0)  and  (0,  —  2)  is  always  equal  to  8.     Find  the  equation  of  the 
locus  and  plot. 

Ans.    The  parallel  straight  lines  6x  +  4j/  +  3  =  0,  6x  +  4y-13=0. 

6.  A  point  moves  so  as  to  be  always  equidistant  from  the  axes  of  coor- 
dinates.    Find  the  equation  of  the  locus  and  plot. 

Ans.    The  perpendicular  straight  lines  x  +  y  =  0,  x  —  y  =  0. 

7.  A  point  moves  so  as  to  be  always  equidistant  from  the  straight  lines 
x  —  4  =  0  and  y  +  5  =  0.     Find  the  equation  of  the  locus  and  plot. 

Ans.   The  perpendicular  straight  lines  x-y-9  =  0,  x  +  y  +  I  =  Q. 

8.  Find  the  equation  of  the  locus  of  a  point  the  sum  of  the  squares  of 
whose  distances  from  (3,  0)  and  (-3,  0)  always  equals  68.     Plot  the  locus. 

Ans.    The  circle  x2  +  y2  =  25. 

9.  Find  the  equation  of  the  locus  of  a  point  which  moves  so  that  its  dis- 
tances from  (8,  0)  and  (2,  0)  are  always  in  a  constant  ratio  equal  to  2.     Plot 
the  locus.  Ans.   The  circle  x2  +  y2  =  16. 

10.  A  point  moves  so  that  the  ratio  of  its  distances  from  (2,  1)  and  (—  4,  2) 
is  always  equal  to  £.     Find  the  equation  of  the  locus  and  plot. 

Ans.   The  circle  3x2  +  32/2-24x-4z/  =  0. 


I  THE  CURVE  AND  THE  EQUATION  59 

In  the  proofs  of  the  following  theorems  the  choice  of  the  axes 
coordinates  is  left  to  the  student,  since  no  mention  is  made 
of  either  coordinates  or  equations  in  the  problem.    In  such  cases 
always  choose  the  axes  in  the  most  convenient  manner  possible.. 

11.  A  point  moves  so  that  the  sum  of  its  distances  from  two  perpendicular 
lines  is  constant.     Show  that  the  locus  is  a  straight  line. 

Hint.  Choosing  the  axes  of  coordinates  to  coincide  with  the  given  lines,  the  equation 
is  x  +  y  =  constant. 

12.  A  point  moves  so  that  the  difference  of  the  squares  of  its  distances 
from  two  fixed  points  is  constant.     Show  that  the  locus  is  a  straight  line. 

Hint.  Draw  XX'  through  the  fixed  points,  and  Y  Y'  through  their  middle  point.  Then 
the  fixed  points  may  he  written  (a,  0),  (-  a,  0),  and  if  the  "constant  difference  "  be  denoted 
by  k,  we  find  for  the  locus  4  ax  =  k  or  4  ax  =  —  k. 

13.  A  point  moves  so  that  the  sum  of  the  squares  of  its  distances  from 
two  fixed  points  is  constant.     Prove  that  the  locus  is  a  circle. 

Hint.    Choose  axes  as  in  problem  12. 

14.  A  point  moves  so  that  the  ratio  of  its  distances  from  two  fixed  points 
is  constant.     Determine  the  nature  of  the  locus. 

Ans.  A  circle  if  the  constant  ratio  is  not  equal  to  unity  and  a  straight 
line  if  it  is. 

The  following  problems  illustrate  the  # 

Theorem.  If  an  equation  can  be  put  in  the  form  of  a  product  of 
variable  factors  equal  to  zero,  the  locus  is  found  by  setting  each  fac- 
tor equal  to  zero  and  plotting  the  locus  of  each  equation  separately* 


15.  Draw  the  locus  of        4  x2  -  9  y*  =  0. 
Solution.    Factoring, 
(1)  (2x 


Then,  by  the  theorem,  the  locus  consists  of  the  straight  lines 

(2)  2x-3y  =  0, 

(3)  2x  +  3y  =  0. 

Proof.    1.    The  coordinates  of  any  point  (xi,  yi)  which  satisfy  (1)  witt 
satisfy  either  (2)  or  (3). 

For  if  (xi,  yi)  satisfies  (1), 
(4) 


60  ANALYTIC  GEOMETRY 

This  product  can  vanish  only  when  one  of  the  factors  is  zero.     Hence 

either 

2xl-8y1  =  0, 

and  therefore  (xi,  ?/i)  satisfies  (2) ; 

or  2  xi  -f  3  yi  =  0, 

and  therefore  (xi,  y\)  satisfies  (3). 

2.  A  point  (xi,  ?/i)  on  either  of  the  lines  defined  by  (2)  and  (3)  will  also 
lie  on  the  locus  of  (I). 

For  if  (xi,  ?/i)  is  on  the  line  2  x  —  3  y  =  0, 
then  (Corollary,  p.  46) 
(5)  2xi-32/i  =  0. 

Hence  the  product  (2  Xi  —  3  y{)  (2  Xi  -f  3  yi)  also  vanishes,  since  by  (5)  the 
first  factor  is  zero,  and  therefore  (xi,  yi)  satisfies  (1). 

Therefore  every  point  on  the  locus  of  (1)  is  also  on  the  locus  of  (2)  and 
(3),  and  conversely.  This  proves  the  theorem  for  this  example.  Q.E.D. 

16.  Show  that  the  locus  of  each  of  the  following  equations  is  a  pair  of 
straight  lines,  and  plot  the  lines. 

(a)  x2  -  y2  =  0.  (j)  3  x2  +  xy  -  2  y*  +  6  x  -  4  y  =  0. 

(b)  9  X*  -  y'2  =  0.  (k)  x2  -  y2  +  x  +  y  =  0. 

(c)  x2  =  9y2.  (1)  x2  -  xy  +  5x  -  5y  =  0. 

(d)  x-  -  4x  -  5  =  0.  (m)  x2  -2xy  +  ?/2  +  6x  -  Gy  =  0. 

(e)  yi-Qy  =  7.  (n)  x2  -  4 y2  +  5x  -f  10 y  =  0. 

(f)  ?/2-5x?/  +  6?/  =  0.  (o)  x2  +  4x?/  +  4?/2  +  5x  +  Wy  +  6  =  0. 

(g)  xy  -  2x2  -  3x  =  0.  (p)  x2  +  Zxy  +  2?/2  +  x  +  ?/  =  0. 

(h)  xy  -  2x  =  0.  (q)  x2  -  4x?/  -  5y2  +  2x  -  10 y  =  0. 

(i)  x?/  =  0.  (r)  3 x2  -  2 x?/  -  ?/  -f  5x  -  5 y  =  0. 

17.  Show  that  the  locus  of  -4.x2  +  Bx  +  C  =  0  is  a  pair  of  parallel  lines,  a 
single  line,  or  that  there  is  no  locus  according  asA^-B2  —  4  AC  is  positive, 
zero,  or  negative. 

18.  Show  that  the  locus  of  ^.x2  -f  Bxy  +  Cy2  =  0  is  a  pair  of  intersecting 
lines,  a  single  line,  or  a  point  according  &s  A  =  B2  —  4  AC  is  positive,  zero, 
or  negative. 

.   31.  Third  fundamental  problem.    Discussion  of  an  equation. 

The  method  explained  of  solving  the  second  fundamental  prob- 
lem gives  no  knowledge  of  the  required  curve  except  that  it 
passes  through  all  the  points  whose  coordinates  are  determined 
as  satisfying  the  given  equation.  Joining  these  points  gives  a 
curve  more  or  less  like  the  exact  locus.  Serious  errors  may  be 


THE  CURVE  AND  THE  EQUATION 


Gl 


made  in  this  way,  however,  since  the  nature  of  the  curve  between 
any  two  successive  points  plotted  is  not  determined.  This  objection 
is  somewhat  obviated  by  determining  before  plotting  certain  prop- 
erties of  the  locus  by  a  discussion  of  the  given  equation  now  to 
be  explained. 

The  nature  and  properties  of  a  locus  depend,  upon  the  form  of 
its  equation,  and  hence  the  steps  of  any  discussion  must  depend 
upon  the  particular  problem.  In  every  case,  however,  the  fol- 
lowing questions  should  be  answered. 

1.  Is  the  curve  a  closed  curve  or  does  it  extend  out  infinitely  far? 

2.  Is  the  curve  symmetrical  with  respect  to  either  axis  or  the 
origin  ? 

The  method  of  deciding  these  questions  is  illustrated  in  the 
following  examples. 

Ex.  1.    Plot  the  locus  of 

(1)  x2  +  42/2  =  16. 
Discuss  the  equation. 

Solution.    First  step.    Solving  for  x, 

(2)  x  =  ±  2  V4  -  2/2. 

Second  step.   Assume  values  of  y  and  compute  x.    This  gives  the  table. 

Third  step.    Plot  the  points  of  the  table. 

Fourth  step.    Draw  a  smooth  curve  through  these  points. 


X 

y 

x 

y 

±4 

0 

±4 

0 

±3.4 

1 

±  3.4 

-  1 

±2.7 

1* 

±2.7 

-H 

0 

2 

0 

-2 

imag. 

3 

imag. 

-3 

if 


3- 


{x. 


Discussion.  1.  Equation  (1)  shows  that  neither  x  nor  y  can  be  indefi- 
nitely great,  since  x2  and  4  y2  are  positive  for  all  real  values  and  their  sum 
must  equal  16.  Therefore  neither  x2  nor  4y2  can  exceed  16.  Hence  the 
curve  is  a  closed  curve. 

A  second  way  of  proving  this  is  the  following : 

From  (2),  the  ordinate  y  cannot  exceed  2  nor  be  less  than  —  2,  since  the 
expression  4  —  y2  beneath  the  radical  must  not  be  negative.  (2)  also  shows 
that  x  has  values  only  from  —  4  to  4  inclusive. 


62 


ANALYTIC  GEOMETRY 


2.  To  determine  the  symmetry  with  respect  to  the  axes  we  proceed  as 
follows : 

The  equation  (1)  contains  no  odd  powers  of  x  or  y ;  hence  it  may  be  writ- 
ten in  any  one  of  the  forms 

(3)  (x)*  +  4  ( -  y)a  =  16,  replacing  (x,  y}  by  (x,  -  y) ; 

(4)  (-  x)2  +  4  (yy  =  16,  replacing  (x,  y)  by  (-  x,  y) ; 

(5)  (-  x)*  +  4  (-  7/)2  =  16,  replacing  (x,  y)  by  (-  x,  -  y). 

The  transformation  of  (1)  into  (3)  corresponds  in  the  figure  to  replacing 
each  point  P(x,  y)  on  the  curve  by  the  point  Q(x,  —  y).  But  the  points  P 
and  Q  are  symmetrical  with  respect  to  XX',  and  (1)  and  (3)  have  the  same 
locus  (Theorem  III,  p.  52).  Hence  the  locus  of  (1)  is  unchanged  if  each  point 
is  changed  to  a  second  point  symmetrical  to  the  first  with  respect  to  XX'. 
Therefore  the  locus  is  symmetrical  with  respect  to  the  axis  of  x.  Similarly 
from  (4),  the  l$cus  is  symmetrical  with  respect  to  the  axis  of  y,  and  from  (5), 
the  locus  is  symmetrical  with  respect  to  the  origin. 

The  locus  is  called  an  ellipse. 

Ex.  2.    Plot  the  locus  of 

(6)  y2_4x  +  i5  =  o. 

Discuss  the  equation. 

Solution.  First  step.  Solve  the  equation  for  x,  since  a  square  root  would 
have  to  be  extracted  if  we  solved  for  y.  This  gives 

(7)  x  =  £0/2  +  15). 


X 

y 

3f 

0 

4 

±1 

4| 
6 

±2 

±3 

7! 
10 

±4 
±5 

12f 

±6 

etc. 

etc. 

w 


-y 


A" 


Second  step.    Assume  values  for  y  and  compute  x. 


THE  CURVE  AND  THE  EQUATION 


63 


Iince  y*  only  appears  in  the  equation,  positive  and  negative  values  of  y 
the  same  value  of  x.     The  calculation  gives  the  table  on  p.  62. 
j?'or  example,  if  y  =  ±  3, 

then  x  =  I  (9  +  15)  =  6,  etc. 

Third  step.    Plot  the  points  of  the  table. 

Fourth  step.    Draw  a  smooth  curve  through  these  points. 

Discussion.  1.  From  (7)  it  is  evident  that  x  increases  as  y  increases. 
Hence  the  curve  extends  out  indefinitely  far  from  both  axes. 

2.  Since  (6)  contains  no  odd  powers  of  y,  the  equation  may  be  written  in 
the  form  <_,)•-  4  <*)  +  16  =  0 

by  replacing  (x,  y)  by  (x,  —  y).     Hence  the  locus  is  symmetrical  with  respect 
to  the  axis  of  x. 

The  curve  is  called  a  parabola. 

Ex.  3.    Plot  the  locus  of  the  equation 
(8)  xy  -  2  y  -  4  =  0. 

Solution.    First  step.    Solving  for  y, 


Second  step.    Compute  y,  assuming  values  for  x. 

When         x  =  2,  y  =  $  =  oo. 

In  such  cases  we  assume  values  differing 
slightly  from  2,  both  less  and  greater,  as  in 
the  table. 

Third  step.    Plot  the  points. 

Fourth  step.  Draw  the  curve  as  in  the 
figure  in  this  case,  the  curve  having  two 
branches. 

1.  From  (9)  it  appears  that  y  diminishes 
and  approaches  zero  as  x  increases  indefi- 
nitely. The  curve  therefore  extends  indefi- 
nitely far  to  the  right  and  left,  approaching 
constantly  the  axis  of  x.  If  we  solve  (8)  for 
x  and  write  the  result  in  the  form 

x  =  2  +  -, 

y 

it  is  evident  that  x  approaches  2  as  y  increases 
indefinitely.     Hence  the  locus  extends  both 
upward  and  downward  indefinitely  far,  approaching  in  each  case  the  line  x  =  2. 


X 

y 

X 

y 

0 

-2 

0 

-2 

1 

-4 

-1 

-1 

If 

-8 

-2 

—  i 

It 

-16 

-4 

-f 

2 

CO 

-5 

—  4 

2.f 

16 

: 

2  1 

8 

-10 

-1 

3 

4 

etc. 

etc. 

4 

2 

5 

f 

6 

1 

12 

0.4 

etc. 

etc. 

64 


ANALYTIC  GEOMETRY 


2.  The  equation  cannot  be  transformed  by  any  one  of  the  three  substitutions 
(x,  y)  into  (x,  -  y), 
(x,  y)  into  (-x,  y), 
(x,  y)  into  (-  x,  -  y), 

without  altering  it  in  such  a  way  that  the  new  equation  will  not  have  the 
same  locus.  The  locus  is  therefore  not  symmetrical  with  respect  to  either 
axis,  nor  with  respect  to  the  origin. 


This  curve  is  called  an  hyperbola. 

Ex.  4.   Draw  the  locus  of  the  equation 
(10)  4y  =  x3. 

Solution.    First  step.    Solving  for  y, 


X 

y 

x 

y 

0 

0 

0 

0 

1 

i 

-1 

-i 

H 

H 

-H 

-H 

2 

2 

-2 

-2 

2* 

3ft 

-2* 

-8ff 

3 

6f 

-3 

-6f 

8* 

lOff 

-3| 

-lOff 

Second  step.  Assume  values  for  x 
and  compute  y.  Values  of  x  must  be 
taken  between  the  integers  in  order  to 
give  points  not  too  far  apart. 

For  example,  if 


=  i 


=J      =  3,  etc. 


THE  CURVE  AND  THE  EQUATION 


65 


Ek 


^ 


Third  step.    Plot  the  points  thus  found. 
Fourth  step.   The  points  determine  the  curve  of  the 
figure. 

Discussion.  1.  From  the  given  equation  (10),  a;  and 
y  increase  simultaneously,  and  therefore  the  curve 
extends  out  indefinitely  from  both  axes. 

2.  In  (10)  there  are  no  even  powers  nor  constant 
term,  so  that  by  changing  signs  the  equation  may  be 
written  in  the  form 


replacing  (x,  y)  by  (-  x,  —  y). 

Hence  the  locus  is  symmetrical  with  respect  to  the 
origin. 

The  locus  is  called  a  cubical  parabola. 


32.  Symmetry.  In  the  above  examples  we  have  assumed  the 
definition : 

If  the  points  of  a  curve  can  be  arranged  in  pairs  which  are 
symmetrical  with  respect  to  an  axis  or  a  point,  then  the  curve 
itself  is  said  to  be  symmetrical  with  respect  to  that  axis  or  point. 

The  method  used  for  testing  an  equation  for  symmetry  of  the 
locus  was  as  follows  :  if  (x,  y)  can  be  replaced  by  (x,  —  y)  through- 
out the  equation  without  affecting  the  locus,  then  if  (a,  b)  is  on 
the  locus,  (a,  —  b)  is  also  on  the  locus,  and  the  points  of  the  latter 
occur  in  pairs  symmetrical  with  respect  to  XX',  etc.  Hence 

Theorem  IV.  If  the  locus  of  an  equation  is  unaffected  by  replacing 
y  by  —  y  throughout  its  equation,  the  locus  is  symmetrical  with 
respect  to  the  axis  of  x. 

If  the  locus  is  unaffected  by  changing  x  to  —  x  throughout  its 
equation,  the  locus  is  symmetrical  with  respect  to  the  axis  of  y. 

If  the  locus  is  unaffected  by  changing  both  x  and  y  to  —  x  and 
—  y  throughout  its  equation,  the  locus  is  symmetrical  with  respect 
to  the  origin. 

These  theorems  may  be  made  to  assume  a  somewhat  different 
form  if  the  equation  is  algebraic  in  x  and  y  (p.  10).  The  locus 
of  an  algebraic  equation  in  the  variables  x  and  y  is  called  an 
algebraic  curve.  Then  from  Theorem  IV  follows 


66  ANALYTIC  GEOMETRY 

Theorem  V.  Symmetry  of  an  algebraic  curve.  If  no  odd  powers 
of  y  occur  in  an  equation,  the  locus  is  symmetrical  with  respect  to 
XX';  if  no  odd  powers  of  x  occur,  the  locus  is  symmetrical  with 
respect  to  YY'.  If  every  term  is  of  even*  degree,  or  every  term  of 
odd  degree,  the  locus  is  symmetrical  with  respect  to  the  origin. 

33.  Further  discussion.  In  this  section  we  treat  of  three  uidre 
questions  which  enter  into  the  discussion  of  an  equation. 

3.  Is  the  origin  on  the  curve  ? 
This  question  is  settled  by 

Theorem  VI.  The  locus  of  an  algebraic  equation  passes  through 
the  origin  when  there  is  no  constant  term  in  the  equation. 

Proof.  The  coordinates  (0,  0)  satisfy  the  equation  when  there 
is  no  constant  term.  Hence  the  origin  lies  on  the  curve  (Corol- 
lary, p.  46).  Q.E.D. 

4.  What  values  of  x  and  y  are  to  be  excluded  ? 
Since  coordinates  are  real  numbers  we  have  the 

Rule  to  determine  all  values  of  x  and  y  which  must  be  excluded. 

First  step.  Solve  the  equation  for  x  in  terms  of  y,  and  from  this 
result  determine  all  values  of  y  for  which  the  computed  value  of  x 
will  be  imaginary.  These  values  ofy  must  be  excluded. 

Second  step.  Solve  the  equation  for  y  in  terms  of  x,  and  from 
this  result  determine  all  values  of  x  for  which  the  computed  value 
of  y  will  be  imaginary.  These  values  of  x  must  be  excluded. 

The  intercepts  of  a  curve  on  the  axis  of  x  are  the  abscissas  of 
the  points  of  intersection  of  the  curve  and  XX'. 

The  intercepts  of  a  curve  on  the  axis  of  y  are  the  ordinates  of 
the  points  of  intersection  of  the  curve  and  YY'. 

Rule  to  find  the  intercepts. 

Substitute  y  =  0  and  solve  for  real  values  of  x.  This  gives  the 
intercepts  on  the  axis  of  x. 

Substitute  x  =  0  and  solve  for  real  values  of  y.  This  gives  the 
intercepts  on  the  axis  of  y. 

*  The  constant  term  must  be  regarded  as  of  even  (zero)  degree. 


THE  CURVE  AND  THE  EQUATION 


67 


The  proof  of  the  rule  follows  at  once  from  the  definitions. 
The  rule  just  given  explains  how  to  answer  the  question: 
5.  What  are  the  intercepts  of  the  locus  ? 

34.  Directions  for  discussing  an  equation.  Given  an  equation, 
;he  following  questions  should  be  answered  in  order  before  plot- 
ing  the  locus. 

1.  Is  the  origin  on  the  locus  ?  (Theorem  VI}. 

2.  Is    the  locus   symmetrical  -with  respect  to  the  axes  or  the 
ifjin?  (Theorems  IV  and  V). 

3.  What  are  the  intercepts?  (Rule,  p.  66). 

4.  What  values  of  x  and  y  must  be  excluded?  (Rule,  p.  66). 

5.  Is  the  curve  closed  or  does  it  pass  off  indefinitely  far?  (§  31, 
>.  61). 

Answering  these  questions  constitutes  what  is  called  a  general 
Iscussion  of  the  given  equation. 

Ex.  1.    Give  a  general  discussion  of  the  equation 
(1)  x2 

Draw  the  locus. 


(04) 


1.  Since  the  equation  contains  no  constant  term,  the  origin  is  on  the  curve. 

2.  The  equation  contains  no  odd  powers  of  x;  hence  the  locus  is  symmet- 
rical with  respect  to  YY'. 

3.  Putting  y  =  0,  we  find  x  =  0,  the  intercept  on  the  axis  of  x.     Putting 
x  =  0,  we  find  y  =  0  and  4,  the  intercepts  on  the  axis  of  y. 

4.  Solving  for  x, 
(2)   " 


68  ANALYTIC  GEOMETRY 

Hence  all  values  of  y  between  0  and  4  must  be  excluded,  since  for  such  a 
value  y2  —  4  y  is  negative. 
Solving  for  y, 

(3)  y  =  2  ±  $  Vx2  +  16. 

Hence  no  value  of  x  is  excluded,  since  x2  +  16  is  always  positive. 

6.  From  (3),  y  increases  as  x  increases,  and  the  curve  extends  out 
indefinitely  far  from  both  axes. 

Plotting  the  locus,  using  (2),  the  curve  is  found  to  be  as  in  the  figure. 
The  curve  is  an  hyperbola. 

PROBLEMS 

1.  Give  a  general  discussion  of  each  of  the  following  equations  and  draw 
the  locus. 

(a)  x2  _  4y  =  o.  (n)  Qy2  _  xs  =  o. 

<  (b)  2/2  -  4x  +  3  =  0.  (o)  92/2  +  x8  =  0. 

(c)  x2  +  4 1/2  _  16  =  0.  (p)  2xy  +  3x  -  4  =  0. 

(d)  9x2  +  y2  _  18  =  0.  (q)  x2  -  xy  +  8  =  0. 

(e)  x2  -  4  ?/2  _  16  =  0.  (r)  &  +  xy  -  4  =  0. 

(f)  x2-4y2  +  16  =  0.  (s)  x2  +  2xy  -3y  =  0. 

(g)  x2  -  2/2  +  4  =  o.  (t)  2xy  -  y*  +  4x  =  0. 
(h)  x2  -  y  +  x  =  0.  (u)  3x2  -  y  +  x  =  0. 

-4(i)  xy  -  4  =  0.  (v)  4y2  -  2x  -  y  =  0. 

x  (j)  9y  +  x3  =  0.  (w)  x2  -  ?/2  +  QX  =  o. 

(k)  4x  -  ys  =  0.  — J  (x)  x2  +  4 y2  -f  8 y  =  0. 

(l)6x-y*  =  0.  (y)  9x2  +  ?/2 +  18x- 6y  =  0. 

(m)  5x  -  ?/  +  2/3  =  0.  (z)  9x2  -  2/2  +  18x  +  6y  =  0. 

2.  Determine  the  general  nature  of  the  locus  in  each  of  the  following 
equations  by  assuming  particular  values  for  the  arbitrary  constants,  but  not 
special  values,  that  is,  values  which  give  the  equation  an  added  peculiarity.* 

(a)  j/2  =  2  mx.  (f )  x2  -  y*  =  a2. 

(b)  x2  -  2  my  =  m2.  (g)  x2  +  y2  =  r2. 
x2      y2  _                                            (h)  x2  +  y2  =  2rx. 

W  ^+52--  (i)  x2  +  2/2  =  2r2/. 

(d)  2xy  =  a2.  (j)  &  -f  y2  =  2ox  +  26y. 
X2      y2                                                 (k)  a2/2  -  a* 

(e)  &-&-  (1)  a^  =  ^3- 

*  For  example,  in  (a)  and  (b)  m=  0  is  a  special  value.  In  fact,  in  all  these  examples 
zero  ia  a  special  value  for  any  constant. 


I  THE  CURVE  AND  THE  EQUATION  69 

3.  Draw  the  locus  of  the  equation 
y»  =  (x-a)(*-6)(x-c), 
(a)  when  a  <  6  <  c.  (c)  when  a  <  6,  b  =  c. 

(b)  when  a  =  6  <  c.  (d)  when  a  =  6  =  c. 

The  loci  of  the  equations  (a)  to  (f  )  in  problem  2  are  all  of  the 
class  known  as  conies,  or  conic  sections  ,  —  curves  following  straight 
lines  and  circles  in  the  matter  of  their  simplicity. 

A  conic  section  is  the  locus  of  a  point  whose  distances  from  a 
fixed  point  and  a  fixed  line  are  in  a  constant  ratio. 

4.  Show  that  every  conic  is  represented  by  an  equation  of  the  second 
degree  in  x  and  y. 

Hint.  Take  Y  Y'  to  coincide  with  the  fixed  line,  and  draw  XX'  through  the  fixed  point. 
Denote  the  fixed  point  by  (p,  0)  and  the  constant  ratio  by  e. 

Ans.  (1  -  e2)x2  +  y2  -  2px  +  p*  =  0. 

5.  Discuss  and  plot  the  locus  of  the  equation  of  problem  4, 

(a)  when  e  =  1.     The  conic  is  now  called  a  parabola  (see  p.  63). 

(b)  when  e  <  1.     The  conic  is  now  called  an  ellipse  (see  p.  62). 

(c)  when  e  >  1.     The  conic  is  now  called  an  hyperbola  (see  p.  64). 

6.  Plot  each  of  the  following. 

a**  -6  =  0.  =  -  ««  =  - 


A    rf'i 

(b)  x*y-y  +  2x  =  0.  (f) 


. 
(c)  xz/2  -  4x  +  6  =  0.  (g)  y  =  ^f?.  (k)  4x  = 


x  +  1  y2  -  9 


35.  Points  of  intersection.  If  two  curves  whose  equations 
are  given  intersect,  the  coordinates  of  each  point  of  intersection 
must  satisfy  both  equations  when  substituted  in  them  for  the 
variables  (Corollary,  p.  46).  In  Algebra  it  is  shown  that  all 
values  satisfying  two  equations  in  two  unknowns  may  be  found 
by  regarding  these  equations  as  simultaneous  in  the  unknowns 
and  solving.  Hence  the 

Rule  to  find  the  points  of  intersection  of  two  curves  whose  equa- 
tions are  given. 


70 


ANALYTIC  GEOMETRY 


First  step.  Consider  the  equations  as  simultaneous  in  the  coordi- 
nates, and  solve  as  in  Algebra. 

Second  step.  Arrange  the  real  solutions  in  corresponding  pairs. 
These  will  be  the  coordinates  of  all  the  points  of  intersection. 

Notice  that  only  real  solutions  correspond  to  common  points 
of  the  two  curves,  since  coordinates  are  always  real  numbers. 

Ex.  1.    Find  the  points  of  intersection  of 

(1)  z-72/  +  25  =  0, 

(2)  z2  +  2/2  =  25. 

Solution.    First  step.    Solving 
(1)  for  x, 

(3)  x  =  1y-  25. 
.      Substituting  in  (2), 

(7 y- 25)2 +  ?/2  =  25. 
Eeducing,  y2  —  7  y  +  12  =  0. 

/.  y  =  3  and  4. 
Substituting  in  (3)  [not  in  (2)], 


x  =  —  4  and  +  3. 


Second  step.  Arranging,  the  points  of  intersection  are  (-4,  3)  and 
(3,  4).  Ans. 

In  the  figure  the  straight  line  (1)  is  the  locus  of  equation  (1),  and  the 
circle  the  locus  of  (2). 

Ex.  2.    Find  the  points  of  intersection  of  the  loci  of 

(4)  2  x2  +  32/2  =  35, 

(5)  3  x2  —  4  y  =  0. 

Solution.    First  step.    Solving  (5)  for  x2, 

(6)  x2  =  f2A 
Substituting  in  (4)  and  reducing, 

9  2/2  -f  8  y  —  105  =  0. 

.-.  y  =  3  and  -  -3/. 

Substituting  in  (6)  and  solving, 
x  =  ±  2  and  ±  i  V-210. 

Second  step.  Arranging  the  real  values,  we  find  the  points  of  intersection 
are  (+2,  3),  (-  2,  3).  Ans. 

In  the  figure  the  ellipse  (4)  is  the  locus  of  (4),  and  the  parabola  (5)  the 
locus  of  (5). 


THE  CURVE  AND  THE  EQUATION  71 

PROBLEMS 

Find  the  points  of  intersection  of  the  following  loci. 


2.  .  Ans.  (6,1). 

- 

'   <0.  «).(-*•  -I)- 


8-    xyt:  20  ^nS<   (±  5'  ±  4)'  (±  4'  ±  5) 

10.  I .    For  what  values  of  6  are  the  curves  tangent  ? 

y  —  o  X  T-  0    J 


G9\    /i          9\ 
ir  =  ^*          j 

x2  =  4  ay        ^1 
13.    y=      8a3       L.  Ans.  (2 a,  a),  (-2a,  a). 


14.      2      9x,  ^Irw.  (8,  6),  (8,  -  6). 

"  2 


(a,  CM-  0,0, 


72  ANALYTIC  GEOMETRY 

X2        V2  X2        V2 

17.  The  two  loci  —  -  --  =  I  and  --  1-  —  =  4  intersect  in  four  points. 

Find  the  lengths  of  the  sides  and  of  the  diagonals  of  the  quadrilateral  formed 
by  these  points. 

Ans.   Points,  (±  VlO,  ±  f  Ve).    Sides,  2  VlO,  3  Ve.    Diagonals,  V94. 

Find  the  area  of  the  triangles  and  polygons  whose  sides  are  the  loci  of  the 
following  equations. 

18.  3x  +  y  +  4  =  0,  3x-5y  +  34  =  0,  3x-2y  +  1  =  0.  Ans.  36. 

19.  x  +  2y  =  5,  2x  +  y  =  7,  y  =  x  +  1.  4ns.  f. 

20.  x  +  y  =  a,  x-2y  =  4a,  y-x  +  7a  =  0.  Ans.  12  a2. 

21.  x  =  0,  y  =  0,  x  =  4,  y  =-6.  4ns.  24. 

22.  x-y  =  0,  x  +  y  =  0,  x-y  =  a,  x  +  y  =  6.  4ns.  —  . 

23.  y  =  3x-9,  y  =  3x  +  6,  2y  =  x  -  6,  2y  =  x  +  14.  4ns.  56. 

24.  Find  the  distance  between  the  points  of  intersection  of  the  curves 
3x-2y  +  6  =  0,  x2  +  y2  =  9.  4ns. 


25.    Does  the  locus  of  y2  =  4x  intersect  the  locus  of  2x  +  3y  +  2  =  0? 

4ns.    Yes. 

26  .  For  what  value  of  a  will  the  three  lines  3x  +  y  —  2  =  0,  ax  +  2y-3  =  0, 
2x  —  y  —  3  =  0  meet  in  a  point  ?  4ns.  a  =  5. 

27.  Find  the  length  of  the  common  chord  of  x2  -f  y2  =  13  and  y2  =  3  x  +  3. 

4ns.   6. 

28.  If  the  equations  of  the  sides  of  a  triangle  are  x  +  7y-fll  =  0, 
3x  +  y  —  7  =  0,  x  —  3y-fl  =  0,  find  the  length  of  each  of  the  medians. 

4ns.    2  V5,  |  V2,  1  VnO. 

Show  that  the  following  loci  intersect  in  two  coincident  points,  that  is,  are 
tangent  to  each  other. 

29.  yt-Wx-  6y-31  =  0,  2y-10x  =  47. 

30.  9x2-4y2  +  54x-16y  +  29  =  0,  15  x  -  8y  +  11  =  0. 

36.  Transcendental  curves.  The  equations  thus  far  consid- 
ered have  been  algebraic  in  x  and  y,  since  powers  alone  of  the 
variables  have  appeared.  We  shall  now  see  how  to  plot  certain 
so-called  transcendental  curves,  in  which  the  variables  appeal- 
otherwise  than  in  powers.  The  Eule,  p.  53,  will  be  followed. 


THE  CURVE  AND  THE  EQUATION 


73 


Ex.  1.    Draw  the  locus  of 

(1)  y  =  logio  x. 

Solution.    Assuming  values  for  x,  y  may  be  computed  by  a  table  of  loga- 
rithms, or,  remembering  the  definition  of  a  logarithm,  from  (1)  will  follow 

(2)  x  =  10". 

Hence  values  may  also  be  assumed  for  y,  and  x  computed  by  (2).     This 
is  done  in  the  table. 
In  plotting, 

unit  length  on  XX'  is  2  divisions, 
unit  length  on  YY'  is  4  divisions. 

General  discussion.  1.  The  curve  does  not 
pass  through  the  origin,  since  (0,  0)  does  not 
satisfy  the  equation. 

2.  The  curve  is  not  symmetrical  with  re- 
spect to  either  axis  or  the  origin. 


X 

y 

X 

y 

1 

0 

.1 

-  1 

3.1 

* 

.01 

-2 

10 

1 

.001 

-3 

100 

2 

.0001 

-4 

etc. 

etc. 

etc. 

etc. 

3.  In  (1),  putting  x  =  0, 

y  =  log  0  =  —  oo  =  intercept  on  YY'. 
In  (2),  putting  y  =  0, 

x  =  10°  =  1  =  intercept  on  XX'. 


4.  From  (2),  since  logarithms  of  negative  numbers  do  not  exist,  all  nega- 
tive values  of  x  are  excluded. 

From  (2)  no  value  of  y  is  excluded. 

5.  From  (2),  as  y  increases  x  increases,  and  the  locus  extends  out  indefi- 
nitely from  both  axes. 

From  (1),  as 

x  approaches  zero, 

y  approaches  negative  infinity  ; 

BO  we  see  that  the  curve  extends  down  indefinitely  and  approaches  nearer 
and  nearer  to  YY'. 


74 


ANALYTIC  GEOMETRY 


Ex.  2.    Draw  the  locus  of 
(3)  y  =  sin  x 

if  the  abscissa  x  is  the  circular  measure  of  an  angle  (Chapter  I,  p.  12). 

Solution.    Assuming  values  for  x  and  finding  the  corresponding  number 
of  degrees,  we  may  compute  y  by  the  table  of  Natural  Sines,  p.  14. 
For  example,  if 

x  =  1,  since  1  radian  =  57°. 29, 

y  =  sin  57°.  29  =  .843.  [by  (3)] 

It  will  be  more  convenient  for  plotting  to  choose  for  x  such  values  that 
the  corresponding  number  of  degrees  is  a  whole  number.  Hence  x  is 
expressed  in  terms  of  it  in  the  table. 

For  example,  if 


X 

y 

x 

y 

0 

0 

0 

0 

Tt 

6 

".50  x 

Tt 

~  6 

-.50 

Tt 

.86 

Tt 

-.86 

8 

3 

Tt 

1.00 

7T 

-1.00 

2 

~2 

27T 

.86 

2tf 

-.86 

3 

3 

5_7r 

.50 

57r 

-.50 

6 

6 

Tt 

0 

-  Tt 

0 

Tt 

x=s' 


=  sin-  =  sin  60°  =  .86. 
3 

2  Tt  2  Tt  9  7T" 

x=  -  — ,  y  =  sin  -  _=  -sin  ~—  (4,  p.  12) 
=  -  sin  120°=  -sin 60° (5,  p.  13) 


In  plotting,  three  divisions  being  taken 
as  the  unit  of  length,  lay  off 

AO-  OB  =  Tt  =  3.1416, 
and  divide  AO  and  OB  up  into  six  equal 
parts. 

The  course  of  the  curve  beyond  B  is 
easily  determined  from  the  relation 

sin  (2  Tt  +  x)  =  sin  x. 
Hence     y  =  sinx  =  sin(2^r  +  x), 

that  is,  the  curve  is  unchanged  if  x  +  2  it  be  substituted  for  x.     This  means, 
however,  that  every  point  is  moved  a  distance  2  Tt  to  the  right.    Hence  the  arc 


Y. 

(0 

,1 

',> 

y 

=7 

V 

^ 

^ 

1 

^N 

s 

^ 

—  --_: 

_-7T 

rl 

7T, 

fl 

"1 

—  . 

F 

~0 

7 

/ 

% 

^ 

* 

i 

\ 

^ 

y^ 

°f/ 

' 

^ 

^ 

\ 

< 

i 

/ 

7T 

j 

7- 

J 

7T 

5J 

r1 

<^  TT 

* 

\ 

\ 

i 

1 

/ 

s 

^ 

K, 

^ 

^ 

/ 

Ss 

^ 

P 

//= 

-1 

^? 

APO  may  be  moved  parallel  to  XX'  until  -4  falls  on  J5,  that  is,  into  the 
position  BBC,  and  ii  will  also  be  a  part  of  the  curve  in  its  new  position. 


THE  CURVE  AND  THE  EQUATION  75 

Also,  the  arc  OQB  may  be  displaced  parallel'  to  XX'  until  O  falls  upon  C. 
In  this  way  it  is  seen  that  the  entire  locus  consists  of  an  indefinite  number 
of  congruent  arcs,  alternately  above  and  below  XX'. 

General  discussion.  1.  The  curve  passes  through  the  origin,  since  (0,  0) 
satisfies  the  equation. 

2.  Since  sin  (—  x)=  —  sin  x,  changing  signs  in  (3), 

—  y  =  —  sin  x, 
or  —  ?/  =  sm(-x). 

Hence  the  locus  is  unchanged  if  (x,  y)  is  replaced  by  (—  x,  —  y),  and  the 
curve  is  symmetrical  with  respect  to  the  origin  (Theorem  IV,  p.  65). 

3.  In  (3),  if  x  =  0, 

y  =  sin  0  =  0  =  intercept  on  the  axis  of  y. 
Solving  (3)  for  x, 

(4)  x  =  sin-  1  y. 

In  (4),  if  y=0, 

x  =  sin-  !  0 
=  nit,  n  being  any  integer. 

Hence  the  curve  cuts  the  axis  of  x  an  indefinite  number  of  times  both 
on  the  right  and  left  of  0,  these  points  being  at  a  distance  of  it  from  one 
another. 

4.  In  (3),  x  may  have  any  value,  since  any  number  is  the  circular  measure 
of  an  angle. 

In  (4),  y  may  have  values  from  —1  to  +1  inclusive,  since  the  sine  of  an 
angle  has  values  only  from  —  1  to  -f  1  inclusive. 

5.  The  curve  extends  out  indefinitely  along  XX'  in  both  directions,  but  is 
contained  entirely  between  the  lines  y  =  +  1,  y  =  —  1. 

The  locus  is  called  the  wave  curve,  from  its  shape,  or  the  sinusoid,  from 
its  equation  (3). 

PROBLEMS 

Plot  the  loci  of  the  following  equations. 

1.  y  =  cosx.  7.    y  =  21ogiox. 

i 

2.  y  =  tan  x.  8.    y  =  (1  +  x)* 

3.  =  secx.  9.    y  =  sin2x. 


4.  y  =  sin-1x.  10.    y  =  tan-- 

5.  y  =  tan~1x.  11.    y  =  2cosx. 

6.  y  =  '2x.  /     12.    y  =  sinx  +  cosx 


CHAPTER  IV 

THE  STRAIGHT  LINE  AND  THE  GENERAL  EQUATION  OF 
THE   FIRST  DEGREE 

37.  The  idea  of  coordinates  and  the  intimate  relation  connect- 
ing a  curve  and  an  equation,  which  results  from  the  introduction 
of  coordinates  into  the  study  of  Geometry,  have  been  considered 
in  the  preceding  chapters.    Analytic  Geometry  has  to  do  largely 
with  a  more  detailed  study  of  particular  curves  and  equations. 
In  this  chapter  we  shall  consider  in  detail  the  straight  line  and 
the  general  equation  of  the  first  degree  in  the  variables  x  and  y 
representing  coordinates. 

38.  The  degree  of  the  equation  of  a  straight  line.   It  was 
shown  in  Chapter  III  (Theorem  I,  p.  51)  that 

(1)  y  =  mm  +  b 

is  the  equation  of  the  straight  line  whose  slope  is  m  and  whose 
intercept  on  the  F-axis  is  b ;  m  and  b  may  have  any  values, 
positive,  negative,  or  zero  (p.  27).  But  if  a  line  is  parallel  to 
the  F-axis,  its  equation  may  not  be  put  in  the  form  (1);  for, 
in  the  first  place,  the  line  has  no  intercept  on  the  F-axis,  and, 
in  the  second  place,  its  slope  is  infinite  and  hence  cannot  be 
substituted  for  m  in  (1).  The  equation  of  a  line  parallel  to  the 
F-axis  is,  however,  of  the  form 

(2)  x  =  constant. 

The  equation  of  any  line  may  be  put  either  in  the  form  (1)  or 
(2).  As  these  equations  are  both  of  the  first  degree  in  x  and  y 
we  have 

Theorem  I.  The  equation  of  any  straight  line  is  of  the  first  degree 
in  the  coordinates  x  and  y. 

76 


THE  STRAIGHT  LINE  77 

39.  The  general  equation  of  the  first  degree,  Avc+By+C=Q. 

The  equation 

(1)  Ax+By  +  C  =  0, 

where  A9  B,  and  C  are  arbitrary  constants  (p.  1),  is  called  the 
general  equation  of  the  first  degree  in  x  and  y  because  every  equa- 
tion of  the  first  degree  may  be  reduced  to  that  form. 
Equation  (1)  represents  all  straight  lines. 

For  the  equation  y  =  mx  +  6  may  be  written  mx  —  y  +  6  =  0,  which  is  of  the 
form  (l)itA  =  m,B  =  —  l,  C=  6;  and  the  equation  x  =  constant  may  be  written 
x  —  constant  =  0,  which  is  of  the  form  (1)  if  A  =  1,  B  =  0,  C  =  —  constant. 

Theorem  II.  (Converse  of  Theorem  I.)  The  locus  of  the  general 
equation  of  the  first  degree 

Ax  +  By  -h  C  =  0 
is  a  straight  line. 

Proof.    Solving  (1)  for  y,  we  obtain 


This  equation  has  the  same  locus  as  (1)  (Theorem  III,  p.  52). 
By  Theorem  I,  p.  51,  the  locus  of  (2)  is  the  straight  line  whose 

A  C 

slope  is  m  =  —  —  and  whose  intercept  on  the  F-axis  is  b  =  —  —  • 
B  B 

If,  however,  B  =  0,  it  is  impossible  to  write  (1)  in  the  form 
(2).  But  if  B  =  0,  (1)  becomes 

Ax  +  C  =  0, 

or 

The  locus  of  this  equation  is  a  straight  line  parallel  to  the 
Faxis  (1,  p.  50).  Hence  in  all  cases  the  locus  of-(i)  is  a  straight 
line.  Q.E.D. 

Corollary  I.    The  slope  of  the  line 

Ax  +  By  +  C  =  0 

is  m  =  —  — ;    that  is,  the  coefficient  of  x  with  its  sign  changed 
B 

divided  by  the  coefficient  of  y. 


78  ANALYTIC  GEOMETRY 

Corollary  II.    The  lines 

Ax  +  By  +  C  =  0 
am?  A  'x  +  £'?/  +  C'  =  0 

are  parallel  when  and  only  when  the  coefficients  of  x  and  y  are 
proportional;  that  is, 

A        B 


For  two  lines  are  parallel  when  and  only  when  their  slopes  are  equal  (Theorem 
VI,  p.  29)  ;  that  is,  when  and  only  when 

_A__A_ 
B  ~      B'' 
Changing  the  signs  and  applying  alternation,  we  obtain 

—  =  — 
A'~  B'' 

Corollary  III.    The  lines 

Ax  +  By  +  C  =  0 

and  A'x  +  B'y  +  C"  =  0 

are  perpendicular  when  and  only  when 

AA'  +  B&  =  Q. 

For  two  lines  are  perpendicular  when  and  only  when  the  slope  of  one  is  the 
negative  reciprocal  of  the  slope  of  the  second  (Theorem  VI,  p.  29)  ;  that  is, 

A  _B' 
~B-^>' 
or  AA'  +  BB'  =  0. 

Corollary  IV.    The  intercepts  of  the  line 
Ax  +  By  -f  C  =  0 
o?i  the  X-  and  Y-axes  are  respectively 

C  C 

a  =  --  -  ana  o  =  —  —  • 

A  .£> 

For  the  intercept  on  the  X-axis  is  found  (p.  66)  by  setting  y  ==  0  and  solving 
for  x,  and  the  intercept  on  the  Y-axis  has  been  found  in  the  above  proof. 

Corollaries  I  and  IV  are  given  chiefly  for  purposes  of  reference.  In  a  numerical 
example  the  intercepts  are  found  most  simply  by  applying  the  general  rule  already 
given  (p.  66)  ;  and  the  slope  is  found  by  reducing  the  equation  to  the  form 

y  =  mx  +  6, 
when  the  coefficient  of  x  will  be  the  slope. 


THE  STRAIGHT  LINE  79 

Theorems  I  and  II  may  be  stated  together  as  follows : 
The  locus  of  an  equation  is  a  straight  line  when  and  only  when 
the  equation  is  of  the  first  degree  in  x  and  y. 

Theorem  II  asserts  that  the  locus  of  every  equation  of  the  first 
degree  is  a  straight  line.  Then,  to  plot  the  locus  of  an  equation 
of  the  first  degree  it  is  merely  necessary  to  plot  two  points  on  the 
locus  and  draw  the  straight  line  passing  through  them.  The  two 
simplest  points  to  plot  are  those  at  which  the  line  crosses  the 
axes.  But  if  those  points  are  very  near  the  origin  it  is  better  to 
use  but  one  of  them  and  some  other  point  not  near  the  origin 
whose  coordinates  are  found  by  the  Rule  on  p.  53. 

Theorem  III.    When  two  equations  of  the  first  degree, 

(3)  Ax  +  By  +  C  =  0 
and 

(4)  A'x  +  B'y  -\-  C1  =  0, 

have  the  same  locus,  then  the  corresponding  coefficients  are  propor- 
tional; that  is, 

A  \  T>f  /~v  I 

Proof.  The  lines  whose  equations  are  (3)  and  (4)  are  by 
hypothesis  identical  and  hence  they  have  the  same  slope  and  the 
same  intercept  on  the  F-axis.  Since  they  have  the  same  slope, 

-  =  —  >  (Corollary  I,  p.  77) 

and  since  they  have  the  same  intercept  on  the  F-axis, 

I  =  j,>  (CoroUary  IV,  p.  78) 

by  alternation  we  obtain 

A        B 


AB 

and  hence 


80  ANALYTIC  GEOMETRY 

Ex.  1.    Find  the  values  of  a  and  b  for  which  the  equations 

2  ox  +  2y  —  5  =  0 

and  4z-3?/  +  7&  =  0 

will  represent  the  same  straight  fine. 

Solution,    These  two  equations  will  represent  the  same  straight  line  if 
(Theorem  III)  .    </a        2         -  5 


and  hence  the  required  values  are  obtained  by  solving 
for  a  and  6.     This  gives 


2a        2  2-5 

_  —  _          an/i          _  —  _ 

T~^3  -3~7&~ 


40.  Geometric  interpretation  of  the  solution  of  two  equations 
of  the  first  degree.  If  we  solve  the  equations 

(1)  Ax  +  By  +  C  =  0 
and 

(2)  A'x  +  B'y  +  C'  =  0, 

we  obtain  the  coordinates  of  the  points  of  intersection  of  the 
lines  whose  equations  are  (1)  and  (2)  (Rule,  p.  69).  But  if 
these  lines  are  parallel  they  do  not  intersect,  and  if  they  are 
identical  they  intersect  in  all  of  their  points.  The  relation 
between  the  position  of  the  lines  whose  equations  are  (1)  and 
(2)  and  the  number  of  solutions  of  the  simultaneous  equations 
(1)  and  (2)  may  be  indicated  as  follows  : 

Number  of  solutions 
Position  of  lines  ft 

of  equations 

Intersecting  lines.  One  solution. 

Parallel  lines.  No  solution. 

Coincident  lines.  An  infinite  number. 

It  is  sometimes  as  convenient  to  be  able  to  determine  the 
number  of  solutions  of  two  equations  of  the  first  degree  without 
solving  them  as  it  is  to  be  able  to  determine  the  nature  of  the 
roots  of  a  quadratic  equation  without  solving  it.  The  following 
theorem  enables  us  to  do  this. 


THE  STRAIGHT  LINE  81 


Theorem  IV.    Two  equations  of  the  first  degree, 

Ax  -f  By  +  C  =  0 

and  A'x  +  B'y  +  C' =  0, 

Aave,  tVi  general,  one  solution  for  x  and  y ;  but  if 

T>  =  3'' 

there  is  no  solution  unless 

A  T>  /~y 

when  there  is  an  infinite  number  of  solutions. 

The  proof  follows  at  once  from  Corollary  II,  p.  78,  and  Theorem  III. 

PROBLEMS 

1 .  Find  the  intercepts  of  the  following  lines  and  plot  the  lines. 

(a)  2  x  +  3  y  =  6.  Ans.  3,  2. 

(b)  7;  +  7  =  1-     -*  Y+  !f   '    ^  ^**  2»  4- 


x  v      *  y 

->\  ~_ — i        / 

"'   q       ^  ~ 
o        0 

(d)  T  +  -^:  =  !-  ^ns-  4»  -  2- 


(c)  ±  -  |  =  1.  ^ns.  3,  -  5. 

o        0 


2.  Plot  the  following  lines. 

(a)  2z-32/  +  5  =  0.  (c)  ?  +  |  =  1. 

J       o 

(b)  ?y-  5-4z  =  0.  (d)  |-|=-1- 

3.  Find  the  equations,  and  reduce  them  to  the  general  form,  of  the  lines 
for  which 

(a)  TO  =  2,  b  =  -  3.  Ans.  2 x  —  y  —  3  =  0. 

(b)  w  =-.5,6=  f.  -4ns.  z  +  2y-3  =  0. 

(c)  m  =  1 ,  6  =  -  f .  .4ns.  4  x  -  10  y  —  25  =  0. 

(d)  a  =  -,  6  =  -  2.  ^ns.  x  -  y  -  2  =  0. 

4 

O  ,_* 

(e)  or  =  ^— ,  6  =  3.  -4ns.  x  +  y  -  3  =  0. 
Hint.    Substitute  in  ?/  =  mx  +  b. 


82  ANALYTIC  GEOMETRY 

4.  Find  the  number  of  solutions  of  the  following  pairs  of  equations  and 
plot  the  loci  of  the  equations. 

<a>  {4  x  +  6  1  +  9  =  Q!  An8'  N°  S0luti°n- 

(b)  (X  ~  y  =  }'  Ans.  One. 
'  \x  +  y  =  l. 

(c)  -I  ~-4  ^)IS'  An  infinite  number. 

20  =  0. 

.  Nosol«tlon. 


5.  Plot  the  lines  2x-3y  +  6  =  0  and  x  -  y  =  0.     Also  plot  the  locus  of 
(2x-3y  +  6)  +  k(x  -  y)  =  0  for  k  =  0,  ±1,  ±2. 

6.  Select  pairs  of  parallel  and  perpendicular  lines  from  the  following. 


4. 


(b)  -  i2  :  8x  +  y  +  1  =  0.  Ans.  LI 

[L3:9x-3y  +  2  =  0. 


(c)  •  i2  :  5  y  +  2  x  =  8.  -4ns.  L2  JL  L3. 


7.  Show   that    the    quadrilateral   whose    sides   are    2z  —  3y  +  4  =  0, 

3x-y-2  =  0,  4x-6yj-9  =  0,  and  6a  —  2y  +  4  =  0isa  parallelogram. 

•  i 

8.  Find  the  equation  of  the  line  whose  slope  is  —  2  which  passes  through 

the  point  of  intersection  of  y  =  3x  +  4  and  y  =  —  x  +  4. 

-4ns.  2x  +  y  —  4  =  0. 

9.  What  is  the  locus  of  y  =  mx  +  6  if  6  is  constant  and  m  arbitrary  ?  if 
m  is  constant  and  6  arbitrary  ? 

10.  Write  an  equation  which  will  represent  all  lines  parallel  to  the  line 

(a)  y  =  2x  +  7.  (c)  y  -  3x  -  4  =  0. 

(b)  y  =  -  x  +  9.  (d)  2  y  -  4x  +  3  =  0. 

11.  Write  an  equation  which  will  represent  all  lines  having  the  same 
intercept  on  the  F-axis  as  (a),  (b),  (c),  and  (d)  in  problem  10. 

12.  Find  the  equation  of  the  line  parallel  to2x  —  3y  =  0  whose  intercept 
on  the  F-axis  is  -  2.  Ans.  2x-3y-6  =  0. 

13.  What  is  the  locus  of  Ax  -f  By  +  C  —  0  if  B  and  C  are  constant  and 
A  arbitrary  ?  if  A  and  B  are  constant  and  C  arbitrary  ? 


THE  STRAIGHT  LINE  83 

41.  Straight  lines  determined  by  two  conditions.  In  Ele- 
mentary Geometry  we  have  many  illustrations  of  the  determina- 
tion of  a  straight  line  by  two  conditions.  Thus  two  points 
determine  a  line,  and  through  a  given  point  one  line,  and  only 
one,  can  be  drawn  parallel  to  a  given  line.  Sometimes,  however, 
there  will  be  two  or  more  lines  satisfying  the  two  conditions ; 
thus  through  a  given  point  outside  of  a  circle  we  can  draw  two 
lines  tangent  to  the  circle,  and  four  lines  may  be  drawn  tangent 
to  two  circles  if  they  do  not  intersect. 

Analytically  such  facts  present  themselves  as  follows.  The 
equation  of  any  straight  line  is  of  the  form  (Theorem  II,  p.  77) 
(1)  Ax  +  By  +  C  =  0, 

and  the  line  is  completely  determined  if  the  values  of  two  of  the 
coefficients  A,  B,  and  C  are  known  in  terms  of  the  third. 

For  example,  if  A  •=  2  B  and  C  =  —  3  B,  equation  (1)  becomes 

2Bx  +  By  —  3B  =  0, 
or  2x  +  y  —  3  =  0. 

Any  geometrical  condition  which  the  line  must  satisfy  gives 
rise  to  an  equation  between  one  or  more  of  the  coefficients 
A,  B,  and  C. 

Thus  if  the  line  is  to  pass  through  the  origin,  we  must  have  (7=0  (Theorem  VI, 

p.  66) ;  or  if  the  slope  is  to  be  3,  then  —  —  =  3  (Corollary  I,  p.  77). 

B 

Two  conditions  which  the  line  must  satisfy  will  then  give  rise 
to  two  equations  in  A,  B,  and  C  from  which  the  values  of  two  of 
the  coefficients  may  be  determined  in  terms  of  the  third,  and  the 
line  is  then  determined. 

If  these  equations  are  of  the  first  degree,  there  will  be  only  one 
line  fulfilling  the  given  conditions,  for  two  equations  of  the  first 
degree  have,  in  general,  only  one  solution  (Theorem  IV,  p.  81). 
If  one  equation  is  a  quadratic  and  the  other  of  the  first  degree, 
then  there  will  be  two  lines  fulfilling  the  conditions,  provided 
that  the  solutions  of  the  equations  are  real.  And,  in  general, 
the  number  of  lines  fulfilling  the  two  given  conditions  will 
depend  on  the  degrees  of  the  equations  in  the  A9  ^,'and  C  to 
which  they  give  rise. 


84  ANALYTIC  GEOMETRY 

Rule  to  determine  the  equation  of  a  straight  line  which  satisfies 
two  conditions. 

First  step.    Assume  that  the  equation  of  the  line  is 


Second  step.  Find  two  equations  between  A,  B,  and  C  each  of 
which  expresses  algebraically  the  fact  that  the  line  satisfies  one 
of  the  given  conditions. 

Third  step.  Solve  these  equations  for  two  of  the  coefficients  A, 
B)  and  C  in  terms  of  the  third. 

Fourth  step.  Substitute  the  results  of  the  third  step  in  the  equa- 
tion in  the  first  step  and  divide  out  the  remaining  coefficient.  The 
result  is  the  required  equation. 

Ex.  1.  Find  the  equation  of  the  line  through  the  two  points  PI  (5,  —  1) 
and  P2(2,  -2). 

Solution.    First  step.    Let  the  required  equation  be 

(1)  Ax  +  By  4-  C  =  0. 

Second  step.    Since  PI  lies  on  the  locus 
of  (1)  (Corollary,  p.  46), 

(2)  5  A  —  B  -f  C  =  0; 
and  since  P2  lies  on  the  line, 

(3)  2^-2^+07  =  0. 

Third  step.    Solving  (2)  and  (3)  for  A  and  B  in  terms  of  C,  we  obtain 

A  1C1      7? 3  f1 

A.  =—  £C,   -D  —  -5  O. 

Fourth  step.    Substituting  in  (1), 

-  i  Cx  4-  f  Cy  +  C  =  0. 

Dividing  by  C  and  simplifying,  the  required  equation  is 
x  —  3y  -  8  =  0. 

Ex.  2.  Find  the  equation  of  the  line  passing  through  Pj.  (3,  -  2)  whose 
slope  is  —  i. 

Solution.     First  step.    Let  the  re- 
quired equation  be 

(4)  Ax  +  By  4-  C  =  0. 
Second  step.    Since  PI  lies  on  (4), 

(5)  3^4-2^4-0  =  0; 
and  since  the  slope  is  —  J, 


PCS,- 


(6) 
11 


THE  STRAIGHT  LINE  85 

Third  step.    Solving  (5)  and  (6)  for  4  and  C  in  terms  of  B,  we  obtain 

Fourth  step.    Substituting  in  (4), 


or 


PROBLEMS  *  -        w      $"  ' 

1.  Find  the  equation  of  the  line  satisfying  the  following  conditions  and 
plot  the  lines. 

(a)  Passing  through  (0,  0)  and  (8,  2).  Ans.  x  —  4  y  =  0. 

(b)  Passing  through  (—  1,  1)  and  (-3,  1).  Ans.  y  —  I  =  0. 
Passing  through  (—3,  1)  and  slope  =  2.  Ans.  2x  —  7/  +  7  =  0. 
Having  the  intercepts  a  =  3  and  fr  =  -  2._  Ans.  2  x  —  3  y  -  G  =  0. 

(e)  Slope  =  —  3,  intercept  on  X-axis  =  4.  Ans.  3  x  -f  ?/  —  12  =  0. 

(f  )  Intercepts  a  =  —  3  and  b  =  -  4.  Ans.  4x  +  3y  +  12  =  0. 

(g)  Passing  through  (2,  3)  and  (—  2,  —  3).  Ans.  3x  -  2  y  =  0. 

(h)  Passing  through  (3,  4)  and  (-  4,  —  3).  Ans.  x  —  y  +  1  =  0. 

(i)  Passing  through  (2,  3)  and  slope  =  —  2.  4ns.  2  x  +  y  —  7  =  0. 

(j)  Having  the  intercepts  2  and  —  5.  4ns.   -  —  -  =  1. 

2      5 

2.  Find  the  equation  of  the  line  passing  through  the  origin  parallel  to  the 
line  2  x  —  3  y  —  4.  Ans.   2  x  -  3  y  =  0. 

3.  Find  the  equation  of  the  line  passing  through  the  origin  perpendicular 
to  the  line  5x  +  y  —  2  =  0.  Ans.   x  —  5  y  =  0. 

4.  Find  the  equation  of  the  line  passing  through  the  point  (3,  2)  parallel 
to  the  line  4x-y-3  =  0.  Ans.    4  x  -  y  -  10  =  0. 

5.  Find  the  equation  of  the  line  passing  through  the  point  (3,  0)  perpen- 
dicular to  the  line  2x  +  2/  —  5  =  0.  4ns.    x-2y-3  =  0. 

6.  Find  the  equation  of  the  line  whose  intercept  on  the  Y-axis  is  5  which 
passes  through  the  point  (6,  3).  4ns.    x  +  3  y  —  15  =  0. 

7.  Find  the  equation  of  the  line  whose  intercept  on  the  X-axis  is  3  which 
is  parallel  to  the  line  x-4y  +  2  =  0.  4ns.    x-4y-3  =  0. 

8.  Find  the  equation  of  the  line  passing  through  the  origin  and  through 
the  intersection  of  the  lines  x  —  2y  +  3  =  0  and  x  +  2y  —  9  =  0. 

4ns.    x  —  y  =  0. 

9.  Find  the  equation  of  the  straight  line  whose  slope  is  ra  which  passes 
through  the  point  PI  (xi,  y{).  Ans.   y  —  y\  =  m  (x  —  Xi). 


86  ANALYTIC  GEOMETRY 

10.  Find  the  equation  of  the  straight  line  whose  intercepts  are  a  and  b. 


Ans.      +     =  l. 
a      o 


11.  Find  the  equation  of  the  straight  line  passing  through  the  points 

andP2(x2,  2/2). 

.   (y2  -  ?/i)  x  -  (x2  -  xi)  y  +  z2i/i  -  Xiy2  =  0. 


12.  Show  that  the  result  of  the  last  problem  may  be  put  in  the  form 
x-xi  _  y  -  3/1 

OJ2  ~Xi        2/2  -  Vl 
Hint.    Add  and  subtract  xtylt  factor,  transpose,  and  express  as  a  proportion. 

42.  The  equation  of  the  straight  line  in  terms  of  its  slope 
and  the  coordinates  of  any  point  on  the  line.  In  this  section 
and  in  those  immediately  following,  the  Rule  in  the  preceding 
section  is  applied  to  the  determination  of  general  forms  of  the 
equations  of  straight  lines  satisfying  pairs  of  conditions  which 
occur  frequently.  These  general  forms  will  then  enable  us  to 
write  the  equations  of  certain  straight  lines  with  the  same  ease 
that  the  equation  y  =  mx  +  b  enables  us  to  write  the  equation 
of  the  straight  line  whose  slope  and  intercept  on  the  F-axis  are 
given. 

Theorem  V.  Point-slope  form.  The  equation  of  the  straight  line 
which  passes  through  the  point  Pl  (x^  y^)  and  has  the  slope  m  is 

/"TT\  /  \ 

(\  )  y  —  f/i  =  w(a?  —  g^). 

Proof.    First  step.    Let  the  equation  of  the  given  line  be 

(1)  Ax  +  By  +  C  =  0. 

Second  step.    Then,  "by  hypothesis, 

(2)  Axl  +  %!  +  C  =  0 
and 

(3)  -|  =  m. 

Third  step.  Solving  (2)  and  (3)  for  A  and  C  in  terms  of  B, 
we  obtain 

A  =  —  mB  and  C  =  B  (mxl  —  y-^). 


THE  STRAIGHT  LINE  »  87 

Fourth  step.    Substituting  in  (1),  we  have 

-  mBx  +  By  +  B(mxl  —  yi)  =  0. 
Dividing  by  B  and  transposing, 

y  —  2/i  =  m(x  —  «i).  Q.E.D. 

If  P1  lies  on  the  F-axis,  xl  —  0  and  yl  =  b,  so  that  this  equa- 
tion becomes  y  =  mx  -f  b. 

43.  The  equation  of  the  straight  line  in  terms  of  its  intercepts. 

We  pass  now  to  the  consideration  of  a  line  determined  by  two 
points,  and  we  consider  first  the  case  in  which  the  two  points  lie 
on  the  axes.  This  section  does  not,  therefore,  apply  to  lines  par- 
allel to  one  of  the  axes  or  to  lines  passing  through  the  origin,  as  in 
the  latter  case  the  two  points  coincide  and  hence  do  not  deter- 
mine a  line. 

Theorem  VI.    Intercept  form.    If  a  and  b  are  the  intercepts  of  a  line 
on  the  X-  and  Y-axes  respectively,  then  the  equation  of  the  line  is 


Proof.     First  step.    Let  the  equation  of  the  given  line  be 

(1)  Ax  +  By  +  C  =  0. 

Second  step.    By  definition  of  the  intercepts  (p.  66),  the  points 
(a,  0)  and  (0,  b)  lie  on  the  line;  hence 

(2)  Aa  +  C  =  0, 

(3)  Bb  +  C  =  0. 

Third  step.    Solving  (2)  and  (3)  for  A  and  B  in  terms  of  C, 
we  obtain  . 

A  = C  and  B  =  —  -  C. 

<i  o 

Fourth  step.    Substituting  in  (1),  we  have 

-  -  Cx  -  \  Cy  - 
a  b 

Dividing  by  C  and  transposing, 


-  -  Cx  -  r  Cy  +  C  =  0. 
a  b 


a 


88 


ANALYTIC  GEOMETRY 


Ex.  1 .    Write  the  equation  of  the  locus  of  2x  —  6  y  +  3  =  0  in  terms  of 
its  intercepts  and  plot  the  line. 

Solution.     Transposing  the  constant  term,  we  have 


Dividing  by  —  3, 


This  equation  is  of  the  form  (VI).     Hence 
a  =  —  f  and  b  =  £. 
Plotting  the  points  (-  f,  0)  and  (0,  £)  and  joining  them  by  a  straight  line, 
we  have  the  required  line. 

44.  The  equation  of  the  straight  line  passing  through  two 
given  points. 

""  Theorem  VII.    Two-point  form.    The  equation  of  the  straight  line 
passing  through  Pl  (x^  y-^)  and  P2  (x2,  y2)  is 

(Vii)  a?-a?i  =  y-vi 

«2-«i       Vz-Vi 
Proof.    Let  the  equation  of  the  line  be 

(1)  Ax  +  By  +  C  =  0. 
Then,  by  hypothesis, 

(2)  AXl  +  Byi  +  C  =  0 
and 

(3)  Ax2  +  By2  +  C  =  0. 

To  follow  the  Rule,  p.  84,  we  must  solve  (2)  and  (3)  for  A 
and  B  in  terms  of  C,  substitute  in  (1),  and  divide  by  C ;  that  pro- 
cedure amounts  to  eliminating  A,  J5,  and  C  from  (1),  (2),  and  (3), 
and  that  elimmation  may  be  more  conveniently  performed  as 
follows : 

Subtract  (2)  from  (1)  ;  this  gives 

or 

(4)  A(x-x,)  =  - 


THE  STRAIGHT  LINE  89 


Similarly,  subtracting  (2)  from  (3),  we  obtain 

/K\  A  (v    v\- 

(&)  A^XZ  —  X!)- 

Dividing  (4)  by  (5),  we  find 

*C  *~~  3% 

—  •  Q.E.D. 

X2  —  xl      y%  — 

Corollary.  The  condition  that  three  points,  Pl  (a^,  3^),  P2  (x2,  ?/2), 
and  P&  (xS)  y^)  should  lie  on  a  line  is  that 

xz  —  Xi      2/2  —  2/i 

For  this  is  the  condition  that  P3  should  lie  on  the  line  (VII)  passing  through 
Pi  and  P2  (Corollary,  p.  46). 

The  method  of  proving  the  corollary  should  be  remembered 
rather  than  the  corollary  itself,  as  then  the  condition  may  be 
immediately  written  down  from  (VII). 

PROBLEMS 

1.  Find,  by  substitution  in  the  proper  formulas,  the  equations  of  the  lines 
satisfying  the  conditions  in  problem  1,  p.  85. 

2.  Find  the  equations  of  the  lines  fulfilling  the  following  conditions  and 
plot  the  lines. 

Passing  through  the  origin,  slope  =  3.    '         -4ns.  3z  —  y  =  0. 

(b)  Passing  through  (3,  -  2)  and  (0,  —  1).  Ans.  z  +  3y  +  3  =  0. 

(c)  Having  the  intercepts  4  and  —  3. ^V          Ans.  3  z  -  4  y  —  12  =  0. 

(d)  F-intercept  =  5  and  slope  =  3.--  Ans.  3z  —  y  +  5  =  0. 
Passing  through  (1,  -  2)  and  (3,  —  4).  ,  Ans.  z  -f-  y  +  1  =  0. 

(f)  Having  the  intercepts  —  1  and  —  3.  -4ns.  3z  +  y  +  3  =  0. 

(g)  Passing  through  (-  £,  f )  and  slope  =  -  f .      Ans.  4  z  +  6  y  -  7  =  0. 
Passing  through  (0,  0)  and  slope  =  m.  Ans.  y  =  mx. 

3.  Find  the  equations  of  the  sides  of  the  triangle  whose  vertices  are 
(-3,  2),  (3,  -2),  and(0,  - 1). 

Ans.   2z-f-3y  =  0,  z-j-3y-f-3  =  0,  and  z  -|-  y  -f  1  =  0. 

4.  Find  the  equations  of  the  medians  of  the  triangle  in  problem  3  and 
show  that  they  meet  in  a  point. 

Hint.  To  show  that  three  lines  meet  in  a  point,  tind  the  point  of  intersection  of  two 
of  them  and  prove  that  it  lies  on  the  third. 


^ 

'"""'(h) 


90  ANALYTIC  GEOMETRY 

5.  Show  that  the  medians  of  any  triangle  meet  in  a  point. 

Hint.    Taking  one  vertex  for  origin  and  one  side  for  the  X-axis,  the  vertices  may  then 
be  called  (0,  0),  (a,  0),  and  (b,  c). 

6.  Determine  whether  or  not  the  following  sets  of  points  lie  on  a  straight 
line. 

(a)  (0,  0),  (1,  1),  (7,  7).  Am.  Yes. 

(b)  (2,  3,),  (-  4,  -  6),  (8,  12).  Ans.  Yes. 

(c)  (3,  4),  (1,  2),  (5,  1).  Ans.  No. 

(d)  (3,  -  1),  (-  6,  2),  (-  I,  1).  Ans.  No. 

(e)  (5,  6),  (jf,  1),  (-  1,  -  |).  Ana.  Yes. 

(f)  (7,  6),  (2,  1),  (6,  -  2).  Ans.  No. 

7.  Reduce  the  following  equations  to  the  form  (VI)  and  plot  their  loci. 

(a)  2 x  +  3 y  -  6  =  0.  (d)  3x  +  4 y  +  1  =  0. 

(b)  x  -  3  y  +  6  =  0.  (e)  2  x  -  4  y  -  7  =  0. 

(c)  3x  -  4 y  +  9  =  0.  (f)  7 x  -  6 y  -  3  =  0. 

8.  Find  the  equations  of  the  lines  joining  the  middle  points  of  the  sides 
of  the  triangle  in  problem  3  and  show  that  they  are  parallel  to  the  sides. 

Ans.   4x  +  6y  +  3  =  0,  x  -f  3  y  =  0,  and  x  +  y  =  0. 

9.  Find  the  equation  of  the  line  passing  through  the  origin  and  through 
the  intersection  of  the  lines  x  +  2 y  =  1  and  2x  —  4y  —  3  =  0. 

Ans.   x  +  10  y  =  0. 

10.  Show  that  the  diagonals  of  a  square  are  perpendicular. 
Hint.    Take  two  sides  for  the  axes  and  let  the  length  of  a  side  be  a. 

11.  Show  that  the  line  joining  the  middle  points  of  two  sides  of  a  triangle 
is  parallel  to  the  third. 

Hint.    Choose  the  axes  so  that  the  vertices  are  (0,  0),  (a,  0),  and  (b,  c). 

12.  Find  the  equation  of  the  line  passing  through  the  point  (3,  —  4)  which 
has  the  same  slope  as  the  line  2  x  —  y  =  3.  Ans.   2x  —  y  —  10  =  0. 

13.  Find  the  equation  of  the  line  passing  through  the  point  (—  1,  4)  which 
is  parallel  to  the  line  3x  +  y  +  1  =  0.  Ans.    3x  +  y-l  =  0. 

14.  Two  sides  of  a  parallelogram  are2x  +  3y-7  =  0  and  x-3y  +  4  =  0. 
Find  the  other  two  sides  if  one  vertex  is  the  point  (3,  2). 

Ans.   2x  +  37/-12  =  0  and  x  -3y  +  3  =  0. 

15.  Find  the  equation  of  the  line  passing  through  the  point  (—2,  3) 
which  is  perpendicular  to  the  line  x  +  2 y  =  1.       Ans.    2x  —  ?/  +  7  =  0. 


I  THE  STRAIGHT  LINE  91 

16.  Show  that  the  three  lines  x  —  2  y  =  0,  x  +  2y-8  =  0,  and  x  +  2 y 
1  +  k(x  —  2  y)  =  0  meet  in  a  point  no  matter  what  value  k  has. 


17.  Derive  (V)  and  (VII)  by  the  Rule  on  p.  46,  using  Theorem  V,  p.  28. 

18.  Derive  (VI)  and  (VII)  by  the  Rule  on  p.  46,  using  the  theorem  that 
corresponding  sides  of  similar  triangles  are  proportional. 


19.  Derive  y  =  mx  +  b  and  (V)  by  the  Rule  on  p.  46,  using  the  definition 
of  the  tangent  of  an  acute  angle  in  a  right  triangle. 

20.  Derive  the  equation  of  the  straight  line  in  terms  of  the  perpendicular 

distance  p  from  the  origin  to  the  line  and  the  angle  w  which 
that  perpendicular  makes  with  the  positive  direction  of 
the  X-axis. 

Hint.    Find  the  intercepts  in  terms  of  p  and  w  by  solving  the 
right  triangles  in  the  figure  and  substitute  in  (VI). 

X  Ans.   x  cos  w  +  y  sin  w  —  p  =  0. 


0 


21.  What  is  the  locus  of  (V)  if  x\  and  yi  are  constant  and  m  arbitrary  ? 

22.  What  is  the  locus  of  (VI)  if  a  is  constant  and  b  arbitrary  ?  if  b  is  con- 
stant and  a  arbitrary  ? 

23.  Write  an  equation  which  represents  all  lines  passing  through  (2,  —  1). 

24.  Write  an  equation  representing  all  lines  whose  intercept  on  the  X-axis 
is  3. 

25.  Write  in  two  different  forms  the  equation  of  all  lines  whose  intercept 
on  the  F-axis  is  —  2. 

26.  Write  an  equation  representing  all  lines  whose  slope  is  —  £. 

27.  If  the  axes  are  oblique  and  make  an  angle  of  w,  then  the  equation  of  a 
straight  line  in  terms  of  its  inclination  a  and  intercept  on  the  F-axis  b  is 

sin  a 

y  =  -  —  7       -rX  +  b. 
sm  (w  —  a) 

28.  If  the  angle  between  the  axes  is  w,  the  equation  of  the  line  passing 
through  PI(XI,  y\)  whose  inclination  is  a  is 

sin  a 


29.  Show  that  equations  (VI)  and  (VII)  hold  for  oblique  coordinates. 


92 


ANALYTIC  GEOMETRY 


45.  The  normal  form  of  the  equation  of  the  straight  line. 

In  the  preceding  sections  the  lines  considered  were  determiiied 
by  two  points  or  by  a  point  and  a  direction.  Both  of  these 
methods  of  determining  a  line  are  frequently  used  in  Elementary 
Geometry,  but  we  have  now  to  consider  a  line  as  determined  by 
two  conditions  which  belong  essentially  to  Analytic  Geometry. 


B    X 


Let  AB  be  any  line,  and  let  ON  be  drawn  from  the  origin  perpen- 
dicular to  AB  at  C.  Let  the  positive  direction  on  ON  be  from  O 
toward  N,  —  that  is,  from  the  origin  toward  the  line,  —  and  denote 
the  positive  directed  length  OC  by  p  and  the  positive  angle 
XON,  measured,  as  in  Trigonometry  (p.  11),  from  OX  as  initial 
line  to  ON  as  terminal  line,  by  <o.*  Then  it  is  evident  from  the 
figures  that  the  position  of  any  line  is  determined  by  a  pair  of 
values  ofp  and  CD,  both  p  and  CD  being  positive  and  CD  <  2  TT. 

On  the  other  hand,  every  line  determines   a  single  positive 
value  of  p  and  a  single  positive  value  of  CD  which  is  less  than 


2-7T,  unless  p  =  0.  When  j9=*0,  however,  AB  passes  through 
the  origin,  and  the  rule  given  above  for  the  positive  direction 
on  ON  becomes  meaningless.  From  the  figures  we  see  that  we 
can  choose  for  CD  either  of  the  angles  XON  or  XON'.  When 
p  =  0  we  shall  always  suppose  that  CD  <  TT  and  that  the  positive 
direction  on  ON  is  the  upward  direction. 

*  w  is  not  the  angle  between  the  directed  lines  OX  and  OA",  as  defined  on  p.  21. 


THE  STRAIGHT  LINE  93 

Theorem  VIII.    The  normal  form*  of  the  equation  of  the  straight 
line  is 
(VIII)  oc  cos  <o  -|-  y  sin  0)  —  p  =  O, 

where  p  is  the  perpendicular  distance  or  normal  from  the  origin  to 
the  line  and  <o  is  the  positive  angle  which  that  perpendicular  makes 
with  the  positive  direction  OX  of  the  X-axis  regarded  as  initial 
line. 

Proof.    Let  P(x,  y)  be  any  point  on  the  given  line  AB. 

Then  since  AB  is  perpendicular  to 
ON,  the  projection  of  OP  on  ON  is 
equal  to  p  (definition,  p.  22).  By 
the  second  theorem  of  projection 
(p.  41),  the  projection  of  OP  on  ON 
is  equal  to  the  sum  of  the  projections 


X'  0 

r' 


x 


> x    of  OD  and  DP  on  ON.     Then  the  con- 


B 


dition  that  P  lies  on  AB  is 


(1)  proj.  of  OD  on  ON  +  proj.  of  DP  on  ON  =  p. 
By  the  first  theorem  of  projection  (p.  23)  we  have 

(2)  proj.  of  OD  on  ON  =  OD  cos  <o  =  x  cos  to, 

(3)  proj.  of  DP  on  ON  =  DP  cos  (  ^  —  u  }  =  y  sin  u 


For  the  angle  between  the  directed  lines  DP  and  ON  equals  that  between 
V=--  w. 


Substituting  frojn  (2)  and  (3)  in  (1),  we  obtain 

(    x  cos  w  +  y  sin  <D  —  p  =  0.  Q.E.D. 

To  reduce  a  given  equation 

(4)  Ax  -f  By  +  C  =  0 

£0  tfAe  normal  form,  we  must  determine  <o  and  jt?  so  that  the  locus 
of  (4)  is  identical  with  the  locus  of 

(5)  x  cos  w  4-  y  sin  <o  —  j9  =  0. 

*  The  designation  of  this  equation  is  made  clear  by  the  definition  of  the  normal  in 
Chapter  ix. 


94 


ANALYTIC  GEOMETRY 


Then  we  must   have  corresponding  coefficients   proportional 
(Theorem  III,  p.  79). 

cos  co      sin  co      —  p 
""'     A  B       ''  ~C~' 

Denote  the  common  value  of  these  ratios  by  r ;  then 

(6)  cos  to  =  rA, 

(7)  sin  co  =  rBj  and 

(8)  -P  =  rC. 

To  find  r,  square  (6)  and  (7)  and  add ;  this  gives 

sin2  co  +  cos2  co  =  r2(A2  +  B2). 
But  sin2  co  +  cos2  co  =  1 ; 

and  hence  r2(A2  -\-  B2)  =  1,  or 

(9) 


Equation  (8)  shows  which  sign  of  the  radical  to  use ;  for  since 
p  is  positive,  r  and  C  must  have  opposite  signs,  unless  C  =  0.  If 
C  =  0,  then,  from  (8),  p  =  0,  and  hence  co  <  TT  (p.  92) ;  then  sin  <o 
is  positive,  and  from  (7)  r  and  B  must  have  the  same  signs. 

Substituting  the  value  of  r  from  (9)  in  (6),  (7),  and  (8)  gives 

ABC 

p=- 


COS  co  = 


sin  o>  = 


'A2  +  B2 

which  is  the  normal  form  of  (4).     The  result  of  the  discussion 
may  be  stated  in  the  following 

Rule  to  reduce  Ax  +  By  +  C  =  0  to  the  normal  form. 
,  ^irst  step.    Find  the  numerical  value  of  V A2  +  B2. 
Second  step.    Give  the  result  of  the  first  step  the  sign  opposite  to 
that  of  C,  or.  if  C  =  0;  the  same  sicjn  as  that  of  B^ 

Third 'step.    Divide  the  given  equation  by  the  result  of  the  second 
The  result  is  the  required  equation. 


THE  STRAIGHT  LIXE  95 

The  advantages  of  the  normal  form  of  the  equation  of  the 
straight  line  over  the  other  forms  are  twofold.  In  the  first 
place,  every  line  may  have  its  equation  in  the  normal  form; 
whether  it  is  parallel  to  one  of  the  axes  or  passes  through  the 
origin  is  immaterial.  In  the  second  place,  as  will  be  seen  in  the 
following  section,  it  enables  us  to  find  immediately  the  distance 
from  a  line  to  a  point. 

PROBLEMS 

1.  In  what  quadrant  will  ON  (Fig.,    p.  92)  lie  if  sin  w  and  cos  w  are  both 
positive?   both  negative?   if  sinw  is  positive  and  cosw  negative?   if  sinw 
is  negative  and  cos  w  positive  ? 

2.  Find  the  equations  and  plot  the  lines  for  which 

(a)  w  =  0,  p  —  5.  Ans.   x  =  5. 

(b)  «  =  — ,  p  =  3.  Ans.   y  +  3  =  0. 

(c)w  =  -,  p  =  3.  Ans.    V2x  +  V2y-6  =  0. 

(d)  a,  =  —  ,  p  =  2.   -  Ans.   z  -  V3  y  +  4  =  0. 

3 

(e)  u  =  — ,  p  =  4.  Ans.    V2  z  -  >/2  y  -  8  =  0. 

4 


: 


.  Reduce  the  following  equations  to  the  normal  form  and  find  p  and 

(a)  3  x  +  4  y  —  2  =  0.  Ans.  p  =  f  ,  w  =  cos-1  f  =  sin-1  f  . 

(b)  3  x  —  4  y  —  2  =  0.  Ans.   p  =  f  ,  <a  =  cos-1  f  =  sin-1  (—  f). 

(c)  12  x  —  5  y  —  0. 

(d)  2x  +  52/  +  T  -0. 

Ans.  p  = 


(e)  4x  —  3y  +  1  =  0. 


-4ns.   p  =  0,  ( 

«)  =  COS~1(  — 

7                    g  l( 

'      2      ^ 

+  V29? 

^  _  v^9/ 

Ans.  p  =  £,  o 

i>  =  COS~1(  — 

-  ,   ti3  —  COS-  *  1 

'      4      ^ 

\_V41/ 

4.  Find  the  perpendicular  distance  from  the  origin  to  each  of  the  folio   - 
ing  lines. 

(a)  12  x  +  5  y  -  26  =  0.  -4ns.    2. 

(b)  x  +  y  +  1  =  0.  Ans. 

(c)  3x  -2^-1  =  0.  -4ns. 


96 


ANALYTIC  GEOMETRY 


5.  Derive  (VIII)  when  (a)  -<w<it;  (b)  7r<«<^;  (c)  — 

it  22 

(d)  p  =  0  andO<w<-. 

6.  For  what  values  of  p  and  w  will  the  locus  of  (VIII)  be  parallel  to  the 
X-axis  ?  the  F-axis  ?   pass  through  the  origin  ? 

7.  Find  the  equations  of  the  lines  whose  slopes  equal  —  2,  which  are  at  a 
distance  of  5  from  the  origin. 

Ans.   2V5x  +  V5y -25  =  0  and2V5sc +.V5y  +  25  =  0. 

8.  Find  the  lines  whose  distance  from  the  origin  is  10,  which  pass  through 
the  point  (5,  10).  Ans.    y  =  10  and  4x  +  3  y  =  50. 

9.  What  is  the  locus  of  (VIII)  if  p  is  constant  and  w  arbitrary  ?  if  w  is 
constant  and  p  arbitrary  ? 

10.  Write  an  equation  representing  all  lines  whose  distance  from  the 
origin  is  5. 

46.  The  distance  from  a  line  to  a  point.  The  positive  direction 
on  the  normal  ON  drawn  through  the  origin  perpendicular  to  AB 
(Fig.  1)  is  from  0  to  AB  (p.  92);  and  when  AB  passes  through  0 
(Fig.  2)  the  positive  direction  on  ON  is  the  upward  direction. 


•/i 


_r      o/ 


The  positive  direction  on  ON  is  taken  to  be  the  positive  direction 
on  all  lines  perpendicular  to  AB.  Hence  the  distance  from  the 
line  AB  to  the  point  PI  is  positive  if  PI  and  the  origin  are  on 
opposite  sides  of  AB,  and  negative  if  PI  and  the  origin  are  on  the 
same  side  of  AB.  When  AB  passes  through  the  origin  the  distance 
from  A  B  to  PI  is  positive  if  that  distance  is  in  the  upward  direc- 
tion, and  negative  if  it  is  in  the  downward  direction.  Thus  in  the 
figures  the  distance  from  AB  to  P!  is  positive  and  from  AB  to  P2 
is  negative. 


THE  STRAIGHT  LINE 


97 


Theorem  IX.    The  distance  d  from  the  line 
x  cos  o>  4-  y  sin  o>  —  p  =  0 
fo  the  point  Pi(xly  y-^)  is 
(IX)  d  =  #!  cos  co  +  y±  sin  (•>  —  p. 

Proof.    Let  ^45  be  the  given  line  and  let  ON  be  perpendicular 
to  AB.     By  the  second  theorem  of  projection  (p.  41)  we  have 

proj.  of  OPl  on  ON  =  proj.  of  OD  on  ON  +  proj.  of  DPl  on  ON. 

From  the  figure, 
proj.  of  OPi  on  07V 

=  OE  —p  +  c?. 
By  the   first   theorem   of 
projection  (p.  23), 
proj.  of  OD  on  ON 

=  OD  cos  to  =  a?!  cos  OD, 
proj.  of  DPi  on 

=  DPi  cos  (  -  —  <o 
=      sin  Q>. 


Hence 

and  therefore 

From  this  theorem  we  have  at  once  the 

Rule  to  find  the  perpendicular  distance  from  a 


p  -f  d  =  a^  cos  o>  4-  2/i  sin  a>, 
"  , 
d  =  xl  cos  <o  +  yi  sin  <D  —  p.          Q.E.D. 


line  to  a 


JFirst  step.  Reduce  the  equation  of  the  given  line  to  the  normal 
form  (Rule,  p.  9f). 

Second  step.  Substitute  the  coordinates  of  the  given  point  for 
x  and  y  in  the  left-hand  side  of  the  equation.  The  result  is  the 
required  distance. 

The  sign  of  the  result  will  show  on  which  side  of  the  line  the 
point  lies. 


98 


ANALYTIC  GEOMETRY 


/ 

/ 

/ 

f 

/ 

\ 

<! 

/ 

\: 

\ 

/ 

3' 

X 

s 

(2,. 

x 

o 

-  ^ 

Ex.  1.    Find  the  distance  from  the  line  4x  —  3y  +  15  =  0  to  the  point 
(2,1). 

Solution.    First  step.   Reducing  the  grven  equation 
to  normal  form,  we  have 


Second  step.     Substituting  2  for  x  and  1  for  y, 
we  have 

,  =  -f.,  +  l(l>-,  =  -4. 

What  does  the  negative  sign  mean  ? 


Ex.  2.    Prove  that  the  sum  of  the  distances  from  the  legs  of  an  isosceles 
triangle  to  any  point  in  the  base  is  constant. 

Solution.  Take  the  middle  point  of  the  base  for  origin  and  the  base  itself 
for  the  X-axis.  Then  the  values  of  p  for  the  two  legs  are  equal  and  the  values 
of  w  are  supplementary.  Hence,  if  the  equation 
of  one  leg  in  normal  form  is 

x  cos  u  -f  y  sin  <*  —  p  =  0, 
then  the  equation  of  the*other  leg  is . 

x  cos  (it  —  w)  +  y  sin  (it  —  w)  —  p  =  0, 
or  —  x  cos  w  -f  y  sin  w  —  p  =  0. 


JT'  /  "O 
/       Y' 


Let  (a,  0)  be  any  point  in  the  base.  Then  the  distances  from  the  legs  to 
(a,  0)  are  respectively  a  cos  a>  —  p  and  —  a  cos  w  —  j),  so  that  the  sum  of  these 
distances  is  —  2  p,  that  is,  a  constant. 


PROBLEMS 

1.  Find  the  distance  from  the  line 

(a)  x  cos  45°  +  y  sin  45°  -  V2  =  0  to  (5,  —  7). 

(b)  fx-$y-l  =  0  to  (2,1). 

(c)  3£  +  4y  +  15  =  0  to  (-2,  3). 

(d)  2  x  -  7  y  +  8  =  0  to  (3,  -  5). 

(e)  x  -  3  y  =  0  to  <0,  4). 


.   -2V2. 
Ans.   -f. 
. 

49 


J.ns.   — 


V53 


-4ns. 


12 


VTo 


2.  Do  the  origin  and  the  point  (3,  —  2)  lie  on  the  same  side  of  the  line 

.   Yes. 


THE  STRAIGHT  LINE 


3.  Does  the  line  2x  +  3y  +  2  =  0  pass  between  the  origin  and  the  point 
(-2,  3)?  Ans.   No. 

4.  Find  the  lengths  of  the  altitudes  of  the  triangle  formed  by  the  lines 
2  z  +  3  y  =  0,  x  +  3y  +  3  =  0,  and  x  +  y  +  1  =  0. 

Ans.    -4=5i  -  1  and  V£. 
V13     V10  ' 

5.  Find  the  distance  from   the  line  Ax  +  By  +  C  =  0  to  the  point 


6.  Prove  Theorem  IX  when 


.         . 

7.  Find  the  locus  of  all  points  which  are  equally  distant  from 

3x-4y  +  l  =  0and4x  +  3y  -1  =  0.. 

Ans.    7  x  —  y  =  0  and  x  +  ly  —  2  =  0. 

8.  Find  the  locus  of  all  points  which  are  twjpe  as  far  from  the  line 
12x-f6y—  .  1  =  0  as  from  the  Y-axis.  Ana.   14  x  —  5y  +  l  =  0. 

9.  Find  the  locus  of  points  which  are  k  times  as  far  from  4x  —  3  y  +  1  =  0 
/as  from  5  x  -  12  y  =  0.  Ans.   (52  —  25  k)  x  —  (39  -  60  k)  y  +  13  =  0. 

10.  Find  the  bisectors  of  the  angles  formed  by  the  lines  in  problem  9. 

Ans.   77x-99y  +  13  =  0  and  27x  +  21  y  +  13  =  0. 

11.  Find  the  distance  between  the  parallel  lines, 

8 


; 


12.  Derive  the  normal  equation  of  the  line  by  means  of  Theorem  IX. 

13.  Prove  that  the  altitudes  on  the  legs  of  an  isosceles  triangle  are  equal. 

14.  Prove  that  the  three  altitudes  of  an  equilateral  triangle  are  equal. 

15.  Prove  that  the  sum  of  the  distances  from  the  sides  of  an  equilateral 
triangle  to  any  point  is  constant. 

Hint.    Take  the  center  of  the  triangle  for  origin,  with  the  JT-axis  parallel  to  one  aide. 


100  ANALYTIC  GEOMETRY 

\   16.  Find  the  areas  of  the  triangles  formed  by  the  following  lines. 

(a)  2  x  -  3  y  +  30  =  0,  x  =  0,  x  +  V  =  0.  Ans.  30. 

+  y  =  2,  3x  +  4y  -  12  =  0,  x  -  y  +  6  =  0.  ^4ns.  li. 

(c)  3x-4y  +  12  =  0,  x-3?/4-  6  =  0,  2x-y  =  0.  ^4ns.  3?. 

(d)  z  +  3y-3  =  0,  5  x  -  y  -  15  =  0,  x  -  y  +  1  =  0.  ^ins.  8. 

17.  Plot  the  following  lines  and  find  the  area  of  the  quadrilaterals  of 
which  they  are  the  sides. 

(a)  x  =  y,  y  =  6,  x  +  y  =  0,  3  x  +  2  y  -  6  =  0.  Ans.    16f . 

(b)  x  +  2  y  -  5  =  0,  y  =  0,  x  +  4y  +  5  =  0,  2x  +  y-4  =  0.    ^.ns.    18. 

(c)  2  «  -  4  y  +  8  =  0,  x  +  y  =  0,  2x-y-4  =  0,  2x  +  y-3  =  0. 

Ans. 


47.  The  angle  which  a  line  makes  with  a  second  line.    The 

angle  between  two  directed  lines  has  been  defined  (p.  21)  as  the 
angle  between  their  positive  directions.  When  a  line  is  given 
by  means  of  its  equation,  no  positive  direction  along  the  line  is 
fixed.  In  order  to  distinguish  between  the  two  pairs  of  equal 
angles  which  two  intersecting  lines  make  with  each  other  we 
define  the  angle  which  a  line  makes  with  a 
second  line  to  be  the  positive  angle  (p.  11) 
from  the  second  line  to  the  first  line. 

Thus  the  angle  which  L±  makes  with  L2 
is  the  angle  0.     We  speak  always  of  the 

"  angle  which  one  line  makes  with  a  second      ______ J^r~ 2t 

line,"  and  the  use  of  the  phrase  "  the  angle 

between  two  lines  "  should  be  avoided  if  those 

lines  are  not  directed  lines.    We  have  thus  added  a  third  method. 

of  designating  angles  to  those  given  on  p.  11  and  p.  21. 

Theorem  X.    The  angle  9  which  the  line 

L, :  A&  +  Biy  +  d  =  0 
makes  with  the  line 

L2 :  A2x  -f  B2y  +  C2  =  0 
is  given  by 

(X)  tan  *=4?f 


THE  STRAIGHT  LINE- 


icr 


Proof.  Let  a^  and  az  be  the  inclinations  of  Zx  and  £2  respec- 
tively. Then,  since  the  exterior  angle  of  a  triangle  equals  the 
sum  of  the  two  opposite  interior  angles,  we  have 

In  Fig.  1,  (*!  =  0  -f  a2,  or  0  =  «i  —  <**, 

In  Fig.  2,  aa  =  TT  —  0  +  al9  or  0  =  TT  -f  (ax  —  aa). 


'      (2) 


And  since  (5,  p.  13) 

tan  (TT 
we  have,  in  either  case, 


)  =  tan  <£, 

tan  6  =  tan  (o^  —  <*2) 
tan  01  —  tan  a 


tan  a-L  tan 


(by  13,  p.  13) 


But  tan  «!  is  the  slope  of  L1  and  tan  az  is  the  slope  of  Z2 ;  hence 
(Corollary  I,  p.  77) 

_^l     ,     ^2 


14- 


Beducing,  we  get   tan  0  = 


Q.E.D. 


Corollary.     T/*  T/I!  ane?  w2  are  #Ae  slopes  of  two  lines,  then  the 
angle  6  which  the  first  line  makes  with  the  second  is  given  by 


tong= 


102 


ANALYTIC  GEOMETRY 


Ex.  1.    Find  the  angles  of  the  triangle  formed  by  the  lines  whose  equations 


are 


M:Qx  —  y  -  6  =  0, 


Solution.  To  see  which  angles  formed  by  the  given 
lines  are  the  angles  of  the  triangle,  we  plot  the  lines, 
obtaining  the  triangle  ABC.  A  is  the  angle  which 
M  makes  with  i,  so  that  M  takes  the  place  of  LI  in 
Theorem  X  and  L  of  L2. 

Hence 

At  =  6,  J5i=-l; 


Then 


and  hence 


A*  =  2,  B2  =  -  3. 
A2Bl  - 


tan^l  = 


-2  +  18_16 
12  +  3    ~  15 


B  is  the  angle  which  L  makes  with  N,  and  by  Corollary  III,  p.  78,  B  =  — . 
C  is  the  angle  which  N  makes  with  M,  so  that  if 


tanO  = 


^2  =  6,  Bo  =  -  1. 

24  +  6      30      15 
tanC  =  36n  =  32  =  16' 


we  must  set 

Hence 
and 


We  may  verify  these  results.     For  if  B  =  — ,  then  A  = (7;  and  hence 


(6,  p.  13,  and  1,  p.  12)  tan  A  = 
true  for  the  values  found. 


tanC 


is 


Ex.  2.     Find  the   equation   of  the  line   through 

(3,  5)  which  makes  an  angle  of  —  with  the  line 
x  —  y  +  6  =  0. 

Solution.    Let  mi  be  the  slope  of  the  required  line. 
Then  its  equation  is  (Theorem  V,  p.  86) 

(1)  y-BrriMx-S). 


THE  STRAIGHT  LINE  103 


The  slope  of  the  given  line  is  ra2  =  1,  and  since  the  angle  which  (1)  makes 

with  the  given  line  is  —  ,  we  have  (by  the  Corollary), 
3 

it      mi  —  1' 
tan-  =  -      —  , 
3      1  +  ?ni 


1+mi 

whence  mi  =  -  —  =  —  (2  +  V3).. 

1-V3 

Substituting  in  (1),  we  obtain 

y  _  5  =  -  (2  +  V3)  (x  -  3)L 
or  (2  +  V3)x  +  y  -  (11  +  3V3)  =  0. 

In  Plane  Geometry  there  would  be  two  solutions  of  this  problem,  —  the 
line  just  obtained  and  the  dotted  line  of  the  figure.  Why  must  the  latter 
be  excluded  here  ? 

PROBLEMS  ^ 

1.  Find  the  angle.  which  the  line  3x-y+2=0  makes  with  2x+y—  2=0; 
also  the  angle  which  the  second  line  makes  with  the  first,  and  show  that 
these  angles  are  supplementary.  3  it    TC 

ns'   ~7~'  7' 
4       4 

2.  Find  the  angle  which  the  line 

(a)  2  x  —  5  ?/  +  1  =  0  makes  with  the  line  x  —  2  y  +  3  =  0. 

(b)  x  -4-  y  4-  1  =  0  makes  with  the  line  x  —  y  +  1  =  0. 

(c)  3x  —  4y  +  2  =  0  makes  with  the  line  x  +  3y  —  7  =  0. 

(d)  6x-3y  +  3  =  0  makes  with  the  line  x  =  6. 

(e)  x  —  7y  +  l  =  0  makes  with  the  line  x  +  2y  —  4  =  0. 

In  each  case  plot  the  lines  and  mark  the  angle  found  by  a  small  arc. 
Ans.  (a)tan-i(-TV);  (b)£;  (c)tan-i(V);  (d)  tan-i(-  |)  ;  (e)  tan-  *(&). 

A 

3.  Find  the  angles  of  the   triangle  whose   sides   are   x  +  3y  —  4  =  0, 
3x  -  2  y  +  1  =  0,  and  x  -  y  +  3  =  0.  Ans.  tan-^-  V),  tan~i(i),  tan~i(2). 

Hint.  Plot  the  triangle  to  see  which  angles  formed  by  the  given  lines  are  the  angles 
of  the  triangle. 

4  .  Find  the  exterior  angles  of  the  triangle  formed  by  the  lines  5x  —  y+3=0, 
y  =  2,  x-4y  +  3  =  0.  Ans.  tan-!(5),  tan~i(-  £),  tan~!(-  -1/-). 

5.  Find  one  exterior  angle  and  the  two  opposite  interior  angles  of  the 
triangle  formed  by  the  lines  2x-3y-6=0,  3x+4y-12  =  0,  x-3y+6=0. 
Verify  the  results  by  formula  12,  p.  13. 


104  ANALYTIC  GEOMETRY 

6  .  Find  the  angles  of  the  triangle  formed  by3x+2y-4=0,  a-3y+6=0, 
and  4  x  —  3y—  10=0.     Verify  the  results  by  the  formula 

tan  A  +  tan  B  +  tan  C  =  tan  A  tan  B  tan  C,  if  A  +  B  +  C  =  180°. 

7.  Find  the  line  passing  through  the  given  point  and  making  the  given 
angle  with  the  given  line. 

(a)  (2,  1),  -,  2x  -  3y  +  2  =  0.          Ans.  5x  -  y  -  9  =  0. 

4 

(b)  (1,  -  3),  —-,  x  +  2y  +  4  =  0.      4iw.  3x  +  y  =  0. 

(c)  (2,  -  5),  -,  x  +  3y  -  8  =  0.        .4ns.  z  -  2y  -  12  =  0. 

(d)  (xi,  ^i),  <f>,y  =  mx  +  b.  Ans.  y  -  yl  =  -^  —   ^-  (x  -  Xi). 

L  —  m  tan  0 


(e)  (xi,  2/1),  0,  4x  +  By  +  C  =  0.       Ans.  y-yl  =  "      (x  - 


8.  Show  from  a  figure  that  it  is  impossible  to  draw  a  line  through  the  inter- 
section of  two  lines  and  "making  equal  angles  with  those  lines"  in  the 
sense  in  which  we  have  defined  "  the  angle  which  one  line  makes  with  a 
second  line."    Prove  the  same  thing  by  formula  (X).     How  are  the  bisectors 
of  the  angles  of  two  lines  to  be  defined  ? 

9.  Given  two  lines  ii:3x-4y-3  =  0  and  i2:4x-3?/  +  12=:0;  find 
the  equation  of  the  line  passing  through  their  point  of  intersection  such  that 
the  angle  it  makes  with  LI  is  equal  to  the  angle  LZ  makes  with  it. 

Ans.    7x-7y  +  9  =  0. 

48.  Systems  of  straight  lines.  An  equation  of  the  first  degree 
in  x  and  y  which  contains  a  single  arbitrary  constant  will  repre- 
sent an  infinite  number  of  lines,  for  the  locus  of  the  equation 
will  be  a  straight  line  for  any  value  of  the  constant,  and  the  locus 
will  be  different  for  different  values  of  the  constant. 

The  lines  represented  by  an  equation  of  the  first  degree  which 
contains  an  arbitrary  constant  are  said  to  form  a  system.  An 
equation  which  represents  all  of  the  lines  satisfying  a  single  con- 
dition must  contain  an  arbitrary  constant,  for  there  is  an  infinite 
number  of  lines  satisfying  a  single  condition  ;  hence  a  single  geo- 
metrical condition  defines  a  system  of  lines. 

Thus  the  equation  y  =  2x  +  b,  where  6  is  an  arbitrary  constant,  represents  the 
system  of  lines  having  the  slope  2  ;  and  the  equation  y  —  5  =  m  (x  —  3),  where  m 
is  an  arbitrary  constant,  represents  the  system  of  lines  passing  through  (3,  5). 


THE  STRAIGHT  LINE 


105 


Second  rule  to  find  the  equation  of  a  straight  line  satisfying  two 
conditions. 

First  step.  Write  the  equation  of  the  system  of  lines  satisfying 
one  condition. 

Second  step.  Determine  tJie  arbitrary  constant  in  the  equation 
found  in  the  first  step  so  that  the  other  condition  is  satisfied. 

Third  step.  Substitute  the  result  of  the  second  step  in  the  result 
of  the  first  step.  This  gives  the  required  equation. 

This  rule  is,  in  general,  easier  of  application  than  the  rule  on 
p.  84.  It  has  already  been  applied  in  solving  Ex.  2,  p.  102,  and 
will  find  constant  application  in  the  following  sections.  The 
number  of  lines  satisfying  the  conditions  imposed  will  be  the 
number  of  real  values  of  the  arbitrary  constant  obtained  in 
the  second  step. 

Ex.  1.  Find  the  equations  of  the  straight  lines  having  the  slope  f  and 
intersecting  the  circle  x2  -f  y2  =  4  in  but  one  point. 

Solution.    First  step.    The  equation 


represents  the  system  of  lines  whose  slopes  are  f  (Theorem  I,  p.  51). 

Second  step.  The  coordinates  of  the  inter- 
section of  the  line  and  circle  are  found  by  solv- 
ing their  equations  simultaneously  (Rule,  p.  69). 
Substituting  the  value  of  y  in  the  line  in  the 
equation  of  the  circle,  we  have 


or          25x2  +  24 bx  +  (16 ft2  -  64)  =  0. 

The  roots  of  this  equation,  by  hypothesis, 
must  be  equal;  hence  the  discriminant  must 
vanish  (Theorem  II,  p.  3) ;  that  is, 

576&2_100(16&2-64)  =  0, 


whence 


Third  step.    Substitute  these  values  of  6  in  the  equation  of  the  first  step. 
We  thus  obtain  the  two  solutions 


and 


106  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Write  the  equations  of  the  systems  of  lines  defined  by  the  following 
conditions. 

(a)  Passing  through  (-  2,  3). 
fa)  Having  the  slope  —  |.   . 

(c)  Distance  from  the  origin  is  3. 

(d)  Having  the  intercept  on  the  F-axis  =  —  3. 

(e)  Passing  through  (6,  —  1). 

(f )  Having  the  intercept  on  the  X-axis  =  6. 

(g)  Having  the  slope  ^. 

(h)  Having  the  intercept  on  the  F-axis  =  5. 
(i)  Distance  from  the  origin  =  4. 

2.  What  geometric  conditions  define  the  systems  of  lines  represented  by 
the  following  equations  ? 

(a)  2  x  -  3  y  +  4  k  =  0. 

(b)  kx-3y  -1  =  0.      I 

(c)  x  +  y  -  k  =  0. 

(d)  x  +  k  =  0. 

(e)  x  +  2  ky  —  3  =  0. 

(f )  2fcc-3y  +  2  =  0. 

(g)  x  cos  a  +  y  sin  a  +  5  =  0. 

Hint.    Beduce  the  given  equation  to  one  of  the  well-known  forms  of  the  equation  of 
the  first  degree. 

3.  Determine  k  so  that 

(a)  the  line  2x  -3y  +  k  =  Q  passes  through  (-  2,  1).      Ans.  k  =  7. 

(b)  the  line  2  kx  —  5 y  +  3  =  0  has  the  slope  3.  Ans.  k  —  -1/. 

(c)  the  line  x  -f  y  —  k  =  0  passes  through  (3,  4).  Ans.  k  =  7. 

(d)  the  line  3z-4?/  +  A;  =  Ohas  intercept  on  X-axis  —  2. 

-4ns.    k  =  —  6. 

(e)  the  line  x  —  3  ky  +  4  =  0  has  intercept  on  F-axis  =  -  3. 

Ans.    k  =  —  |. 

(f )  the  line  4x  —  3y-f6fc  =  0is  distant  three  units  from  the  origin. 

-4ns.    k  =  ±  f . 

4.  Find  the  equations  of  the  straight  lines  with  the  slope  —  T5^  which  cut 
the  circle  ic2  -f  y2  =  1  in  but  one  point.  Ans.    5x  +  12  y  =  ±  13. 

5.  Find  the  equations  of  the  lines  passing  through  the  point  (1,  2)  which 
cut  the  circle  x2  +  y2  =  4  in  but  one  point.     -4ns.    y  =  2  and  4  z  -f  3  y  =  10. 

6.  Find  the  equation  of  the  straight  line  passing  through  (—2,  5)  which 
makes  an  angle  of  45°  with  the  F-axis.  Ans.  x  -f  y  —  3  =  0. 


THE  STRAIGHT  LINE 


107 


7.  Find  the  equation  of  the  straight  line  which  passes  through  the  point 
(2,  —  1)  and  which  is  at  a  distance  of  two  units  from  the  origin. 

Ans.   x  =  2  and  3  x  —  4  y  =  10. 

8.  Find  the  equation  of  the  straight  line  whose  slope  is  f  such  that  the 
distance  from  the  line  to  the  point  (2,  4)  is  2.  Ans.    3  x  —  4  y  =  0. 

49.  The  system  of  lines  parallel  to  a  given  line. 
Theorem  XI.    The  system  of  lines  parallel  to  a  given  line 

Ax  +  By  +  C  =  0 
is  represented  by 

(XI)  Ax  +  By  +  k  =  O, 

where  k  is  an  arbitrary  constant. 

Proof.  All  of  the  lines  of  the  system  represented  by  (XI)  are 
parallel  to  the  given  line  (Corollary  II,  p.  78).  It  remains  to  be 
shown  that  all  lines  parallel  to  the  given  line  are  represented  by 
(XI).  Any  line  parallel  to  the  given  line  is  determined  by  some 
point  Pl  (XD  y^  through  which  it  passes.  If  PI  lies  on  (XI), 
then  Ax±  -f-  By^  +  k  —  0; 

and  hence  k  =  —  Axl  —  BI/I. 

That  is,  the  value  of  k  may  be  chosen  so  that  the  locus  of  (XI) 
passes  through  any  point"  Pj.  Then  (XI)  represents  all  lines 
parallel  to  the  given  line.  Q.E.D. 

It  should  be  noticed  that  the  coefficients  of  x  and  y  in  (XI) 
are  the  same  as  those  of  the  given  equation. 

Ex.  1.    Find  the  equation  of  the  line  through  the  point  PI  (3,  —  2)  paral- 
lel to  the  line  Li:2x-3?/-4  =  0. 

Solution.    Apply  the  Rule,  p.  105. 
First  step.    The  system  of  lines  parallel  to 
the  given  line  is 


Second   step.     The    required    line    passes 
through  PI;   hence 

2.3-3(-2)  +  fc  =  0, 
and  therefore         k  =  —  12. 

Third  step.     Substituting  this  value  of  k, 
the  required  equation  is 


108 


ANALYTIC  GEOMETRY 


50.  The  system  of  lines  perpendicular  to  a  given  line. 
Theorem  XII.     The  system  of  lines  perpendicular  to  the  given 

Ax  +  By  +  C  =  Q 
is  represented  by 

(XII)  Bx  -  Ay  +  k  =  O, 

where  k  is  an  arbitrary  constant. 

Proof.  All  of  the  lines  of  the  system  represented  by  (XII) 
are  perpendicular  to  the  given  line,  for  (Corollary  III,  p.  78) 
A  R  —  BA  =0.  It  remains  to  be  shown  that  all  lines  perpen- 
dicular to  the  given  line  are  represented  by  (XII).  Any  line 
perpendicular  to  the  given  line  is  determined  by  some  point 
-Pi  (xi,  ?/i)  through  which  it  passes.  If  Pl  lies  on  (XII),  then 


whence 


k  =  A      — 


That  is,  the  value  of  k  may  be  chosen  so  that  the  locus  of  (XII) 
•passes  through  any  point  Px.  Then  (XII)  represents  all  lines 
perpendicular  to  the  given  line.  Q.E.D. 

Notice  that  the  coefficients  of  x  and  y  in  (XII)  are  respectively 
the  coefficients  of  y  and  x  in  the  given  equation  with  the  sign  of 
one  of  them  changed. 

Ex.  1.  Find  the  equation  of  the  line  through  the  point  PI  (—  1,  3)  perpen- 
dicular to  the  line  Zi :  5  <c  —  2  y  +  3  =  0. 

Solution.    Apply  the  Rule,  p.  105. 

First  step.    The  equation  of  the  system  of  lines  perpendicular  to  the  given 
line  is 

2x  +  by  +  k  =  0. 

Second    step.      The    required    line    passes 
through  PI;    hence 

or  k  =  - 13. 

Third  step.    Substitute  this  value  of  k.     The  required  equation  is  then 
2z  +  5y-13=:0. 


YA 

• 

^, 

/ 

L\ 

1\ 

(-1-, 

3  ) 

r 

-^ 

^^ 

( 

"*•* 

^ 

1° 

™ 

THE  STRAIGHT  LINE  109 

PROBLEMS 

1.  Find  the  equation  of  the  straight  line  which  passes  through  the  point 

(a)  (0,  0)  and  is  parallel  to  x  —  3  y  -f  4  =  0.  Ans.  x  —  3  y  =  0. 

(b)  (3,  —  2)  ancMs  parallel  to  x  -f  y  +  2  =  0.          -4ns.  x  +  y  —  1  =  0. 

(c)  (-  5,  6)  and  is  parallel  to2z  +  4y-3  =  0.     Ans.  x  +  2y  —  7  =  0. 

(d)  (—  1,  2)  and  is  perpendicular  to3x  —  4y-t-l  =  0. 

Ans.    4x  +  Sy  —  2  =  0. 

(e)  (-  7,  2)  and  is  perpendicular  tox  —  32/4-4  =  0. 

Ans.    3z  +  y  +  19  =  0. 

2.  Find  the  equations  of  the  lines  drawn  through  the  vertices  of  the 
triangle  whose  vertices  are  (—3,  2),  (3,  —  2),  and  (0,  —  1),  which  are  parallel 
to  the  opposite  sides. 

Ans.  The  sides  of  the  triangle  are 

2  z  +  3  y  =  0,  z  +  3  y  +  3  =  0,  x  +  y  +  1  =  0. 
The  required  equations  are 

2z  +  3y  +  3  =  0,  z  -f  3y  -  3  =  0,  x  +  y  -  1  =  0. 

3.  Find  the  equations  of  the  lines  drawn  through  the  vertices  of  the 
triangle  in  problem  2  which  are  perpendicular  to  the  opposite  sides,  and 
show  that  they  meet  in  a  point. 

Ans.    3z-2y-2  =  0,  3  z  -  y  +  11  =  0,  z  -  y  -  5  =  0. 

4.  Find  the  equations  of  the  perpendicular  bisectors  of  the  sides  of  the 
triangle  in  problem  2,  and  show  that  they  meet  in  a  point. 

Ans.  3  z  -  2  y  =  0,  3z-y-6  =  0,  z  —  y  +  2  =  0. 

5.  The  equations  of  two  sides  of  a  parallelogram  are  3z  —  4y  +  6  =  0  and 
x  +  5  y  —  10  =  0.     Find  the  equations  of  the  other  two  sides  if  one  vertex 
is  the  point  (4,  9).  Ans.    3z  —  4y  +  24  =  0  and  x  +  5  y  -  49  =  0. 

6.  The  vertices  of  a  triangle  are  (2,  1),  (-  2,  3),  and  (4,  -  1).     Find  the 
equations  of  (a)  the  sides  of  the  triangle,  (b)  the  perpendicular  bisectors  of 
the  sides,  and  (c)  the  lines  drawn  through  the  vertices  perpendicular  to  the 
opposite  sides.     Check  the  results  by  showing  that  the  lines  in  (b)  and  (c) 
meet  in  a  point. 

7.  Show  that  the  perpendicular  bisectors  of  the  sides  of  any  triangle  meet 
in  a  point. 

8.  Show  that  the  lines  drawn  through  the  vertices  of  a  triangle  perpen- 
dicular to  the  opposite  sides  meet  in  a  point. 

9.  Find  the  value  of  C  in  terms  of  A  and  B  if  Ax  +  By  -f  C  =  0  passes 
through  a  given  point  PI(XI,  yi);  show  that  the  equation  of  the  system  of 
lines  through  PI  may  be  written  A  (x  —  x\)  +  B  (y  —  y\)  =  0. 


110  ANALYTIC  GEOMETRY 

51.  The  system  of  lines  passing  through  the  intersection  of 
two  given  lines. 

Theorem  XIII.  The  system  of  lines  passing  through  the  intersec- 
tion of  two  given  lines 


and  L2  :  A2x  +  Bzy  +  Cz  =  0 

is  represented  by  the  equation    \ 

(XIII)      ^AtfD  +  Biy  +  Ci  -^  ^U«X  +  J52y  +  Ca)  =  O, 

where  k  is  an  arbitrary  constant. 

Proof.  All  of  the  lines  represented  by  (XIII)  pass  through  the 
intersection  of  Ll  and  L2.  For  let  PI  (a^,  f/x)  be  the  intersection 
of  LI  and  Z2.  Then  (Corollary,  p.  46) 


and  ^^   +  £      +  C   =  0. 


Multiply  the  second  equation  by  k  and  add  to  the  first.     This 
gives 

C2  =  0. 


But  this  is  the  condition  that  P!  lies  on  (XIII). 
That  all  lines  through  the  intersection  of  Ll  and  L2  are  repre- 
sented by  (XIII)  follows  as  in  the  proofs  of  Theorems  XI  and 

XII.  Q.E.D. 

Corollary.    If  LI  and  Lz  are  parallel,  then  (XIII)  represents  the 
system  of  lines  parallel  to  Ll  and  L2. 

For  if  LI  and  LZ  are  parallel,  then 


and  hence 

By  composition, 

AI  B\ 

Hence  LI  and  (XIII)  are  parallel  (Corollary  II,  p.  78). 

Notice  that  (XIII)  is  formed  by  multiplying  the  equation  of 
by  k  and  adding  it  to  the  equation  of  LI. 


THE  STRAIGHT  LINE 


111 


Ex.  1.     Find  the  equation  of  the  line  passing  through  PI  (2,  1)  and  the 
intersection  of  LI  :  3  x  -  5  y  -  10  =  0  and  L2  :  x  +  y  +  1  =  0. 

Solution.    Apply  the  Rule,  p.  105.    The  system  of  lines  passing  through  the 
intersection  of  the  given  lines  is  represented  by 

3x  -  5  y  -  10  -1-  k  (x  +  y  +  1)  =  0. 
If  PI  lies  on  this  line,  then 

6_5-10-4-fc(2  +  l  +  l)  =  0; 
whence  k  =  f . 

Substituting  this  value  of  k  and  simplifying, 
we  have  the  required  equation 


Ex.  2.    Find   the   equation   of   the   line   passing   through   the   intersec- 
tion   of    ii:2x  +  y  +  l  =  0    and    L-2  :  x  —  2  y  -f  1  =  0    and    parallel    to 

Solution.    Apply  the  Rule,  p.  105.     The  equation  of  every  line  through 
the  intersection  of  the  first  two  given  lines  has  the  form 

2x  +  y  +  1  +  k(x  -  2y  +  1)  =  0, 
or     (2  +  k)x  +  (1  -2k)y  +  (1  +  k)  =  0. 

If  this  line  is  parallel  to  the  third  line  (Corollary 
II,  P-  78), 

21     1»         I          O  Z* 
-f"  fC          1  —  £  fC 


K 


4  -3    ' 

whence  k  =  2. 

Substituting  and  simplifying,  we  obtain 


The  geometrical  significance  of  the  value  of  k  in  Theorem  XIII 
is  given  most  simply  when  Ll  and  L2  are  in  normal  form. 

Theorem  XIV.    The  ratio  of  the  distances  from 
LI  :  x  cos  CD!  +  y  sin  ^  —  pi  =  0 
and  L2 :  x  cos  o>2  -f  y  sin  <o2  —  p2  =  0 

£0  a?iy  point  of  the  line 

L  :  x  cos  cox  -j-  y  sin  o^  —  ^  -j-  &  (#  cos  oo2  -f-  y  sin  o>3  —  p2)  =  0 
is  constant  and  equal  to  —  k. 


112  ANALYTIC  GEOMETRY 

Proof.    Let  PI  (xl}  y-^  be  any  point  on  L.     Then 

xl  cos  <DI  -f-  2/x  sin  ^  —  pl  -f  A;  (xj  cos  <o2  +  3/1  sin  o>2  —  jt?2)  =  0, 

x-i  cos  a)!  4-  y\  sin  o>i  —  #1 

and  hence  —  k  =  -  — — «. 

a?!  cos  <o2  +  2/1  sin  a>2  —  ^>2 . 

The  numerator  of  this  fraction  is  the  distance  from  L^  to  Plt 
and  the  denominator  is  the  distance  from  Z2  to  PI  (Theorem  IX, 
p.  97).  Hence  —  k  is  the  ratio  of  the  distances  from  Zt  and  L2 
to  any  point  on  L.  Q.E.D. 

Corollary.  If  k  =  ±  1,  then  L  is  the  bisector  of  one  of  the  angles 
formed  by  LI  and  L2.  That  is,  the  equations  of  the  bisectors  of  the 
angles  between  two  lines  are  found  by  reducing  their  equations  to 
the  normal  form  and  adding  and  subtracting  them. 

For  when  k  =  ±  1  the  numerical  values  of  the  distances  from  L\  and  Lz  to  any 
point  of  L  are  equal. 

The  angle  formed  by  Ll  and  Z2  in  which  the  origin  lies,  or  its 
vertical  angle,  is  called  an  internal  angle  of  Ll  and  L2  5  and  either 
of  the  other  angles  formed  by  Ll  and  L2  is 
called  an  external  angle  of  those  lines.  From 
the  rule  giving  the  sign  of  the  distance  from  a 
line  to  a  point  (p.  96)  it  follows  that  L  lies  in 
the  internal  angles  of  LI  and  L2  when  k  is  nega-  f 

tive,  and  in  the  external  angles  when  k  is  posi-  /, 

tive.     If  the  origin  lies  on  Ll  or  Z2,  the  lines 
must  in  each  case  be  plotted  and  the  angles  in  which  k  is  posi- 
tive found  from  the  figure. 


PROBLEMS 

1.  Find  the  equation  of  the  line  passing  through  the  intersection  of 
2x  —  3  y  +  2  =  0  and  3x  —  4y  —  2  =  0,  without  finding  the  point  of  intersec- 
tion, which 

(a)  passes  through  the  origin. 

(b)  is  parallel  to5«-2y  +  3  =  0. 

(c)  is  perpendicular  to3x  —  2y  +  4  =  0. 

Ans.  (a)  5x-  7y  =  0;  (b)  5x-2y  -50  =  0;  (c)  2z  +  3y  -  58  =  0. 


THE  STRAIGHT  LINE  113 

2.  Find  the  equations  of  the  lirfes  which  pass  flirough  the  vertwtes  of  the 
triangle  formed  by  the  lines  2x-3y  +  l  =  0,  x  -  y  =  0,  and  3x  +  4y-2  =  0 
which  are 

(a)  parallel  to  the  opposite  sides. 

(b)  perpendicular  to  the  opposite  sides. 

Ans.    (a)  3x  +  4y-  7  =  0,  Ux-2ly+    2  =  0,  17  x  -  17  y  +  5  =  0  ; 
(b)  4x'-3y-  1  =.0,  21  x  +  14  y-  10  =  0,  17x  +  17y-9  =  0. 

3.  Find  the  bisectors  of  the  angles  formed  by  the  lines  4x  —  3y  —  1  =  0 
and  3x  —  4y  +  2  =  0,  and  show  that  they  are  perpendicular. 

Ans:  7x  —  1y  +  1  =  0  and  x  +  y  —  3  =  0. 

4.  Find  the  equations  of  the  bisectors  of  the  angles  formed  by  the  lines 
5x  —  12y  +  10  =  0  and  12  x  —  5y  +  15  =  0.    Verify  the  results  by  Theorem  X. 

5.  Find  the  locus  of  a  point  the  ratio  of  whose  distances  from  the  lines 
4x  -3y  +  4  =  0and5x-f  12y-8  =  0isl3to5.       Ans.    9x  +  9y-4  =  0. 

^  6.  Find  the  bisectors  of  the  interior  angles  of  the  triangle  formed  by  the 
lines  4  x  —  3  y  =  12,  5x  —  12  y  —  4  =  0,  and  12  x  -  5  y  —  13  =  0.  Show  that 
they  meet  in  a  point. 

Ans.    7  x  -  9  y  -  16  =  0,  7  x  +  7  y  -  9  =  0,  112  x  -  64  y  -  221  =  0.  • 

7.  Find  the  bisectors  of  the  interior  angles  of  the  triangle  formed  by  the 
lines  5  x  —  12  y  =  0,  5  x  +  12  y  +  60  =  0,  and  12  x  —  5  y  —  60  =  0,  and  show 
that  they  meet  in  a  point. 

Ans.   2  y  +  5  =  0,  17  x  +  7  y  =  0,  17  x  -  17  y  -  60  =  0. 

8.  The   sides  of   a  triangle  are  3x  +  4y-12  =  0,  3x  -  4y  =  0,   and 
4x  +  3y  +  24  =  0.      Show  that  the  bisector  of  the  interior  angle  at  the 
vertex  formed  by  the  first  two  lines  and  the  bisectors  of  the  exterior  angles 
at  the-  other  vertices  meet  in  a  point. 

9.  Find   the  equation  of  the  line  passing   through  the   intersection  of 
x  +  y  —  2  =  0  and  x  —  y  +  6  =  0  and  through  the  intersection  of2x  —  y  +  3  =  0 
and  x  -  3y  +  2  =  0.  Ans.    19x  +  3y  +  26  =  0. 

Hint.    The  systems  of  lines  passing  through  the  points  of  intersection  of  the  two  pairs 

of  lines  are 


and  2ar  —  y  +  3  +  k'(x  —  3y  +  2)=0. 

These  lines  will  coincide  if  (Theorem  III,  p.  79) 

1-fc         —  2  +  6fc 


Letting  p  be  the  common  value  of  these  ratios,  we  obtain 

l  +  fc=2p  +  p*', 
l-k=-P-3pk', 


and 

From  these  equations  we  can  eliminate  the  terms  in  pic*  and  p,  and  thus  find  the  value 
of  k  which  gives  that  line  of  the  flrst  system  which  also  belongs  to  the  second  system. 


114  ANALYTIC  GEOMETRY 

10.  Find  the  equation  of  the  line  passing  through  the  intersection  of 
2x  +  5y  —  3  =  0   and  3x  —  2y  -  1  =  0   and  through  the   intersection  of 
6  =  0.  Ans.    43z  -  35y  -  12  =  0. 


A  figure  composed  of  four  lines  intersecting  in  six  points  is 
called  a  complete  quadrilateral.  The  six  vertices  determine  three 
diagonals  of  which,  two  are  the  diagonals  of  the  ordinary  quadri- 
lateral formed  by  the  four  lines. 

11.  Find  the  equations  of  the  three  diagonals  of  the  complete  quadrilateral 
formed  by  the  lines  x  +  2y  =  0,  3z-4y  +  2  =  0,  x  —  y  +  3  =  0,    and 
3x-2y  +  4  =  0.         Ans.   2x  -  y  +  1  =  0,  x  +  2  =  0,  5x  -  Gy  +  8  =  0. 

12.  Show  that  the  bisectors  of  the  angles  of  any  two  lines  are  perpen- 
dicular. 

13.  Find  a  geometrical  interpretation  of  k  in  (XI)  and  (XII). 

14.  Find  the  geometrical  interpretation  of  k  in  (XIII)  when  LI  and  Lz 
are  not  in  normal  form. 

15.  Show  that  the  bisectors  of  the  interior  angles  of  any  triangle  meet  in 
a  point. 

16.  Show  that  the  bisectors  of  two  exterior  angles  of  a  triangle  and  of  the 
third  interior  angle  meet  in  a  point. 


CHAPTER  V 
THE  CIRCLE  AND  THE  EQUATION  «2+  yz  +  I>3c  +Ey  +  F=  O 

52.  The  general  equation  of  the  circle.  If  (a,  ft)  is  the  center 
of  a  circle  whose  radius  is  r,  then  the  equation  of  the  circle  is 
(Theorem  II,  p.  51) 

(1)  x2  +  y2  -  2  ax  -  2/3y  +  a?  +  ft2  -  r2  =  0, 
or 

(2)  (v-af  +  d, -?)•  =  ,*. 

In  particular,  if  the  center  is  the  origin,  a  =  0,  ft  =  0,  and  (2) 
reduces  to 

(3)  xz  +  y*  =  r\ 
Equation  (1)  is  of  the  form 

(4)  x2  +  if  +  Dx  +  Ey  +  F  =  0, 
where 

(5)  D  =  -  2  a,  E  =  -  2  ft,  and  F  =  a2  +  f?  -  r2. 

Can  we  infer,  conversely,  that  the  locus  of  every  equation 
of.  the  form  (4)  is  a  circle?  By  adding  J  D2  +  \ -E2  to  both 
members,  (4)  becomes 

(6)  (x  +  %Dy+(y+iEy  =  i(D2  +  E2-±F). 

In  (6)  we  distinguish  three  cases: 

If  D2  -f-  E'2  —  4  F  is  positive,  (6)  is  in  the  form  (2),  and  hence 
the  locus  of  (4)  is  a  circle  whose  center  is  (—  J  Z),  —  J  E)  and 
whose  radius  is  r  =  £  Vz>2  +  E2  —  4  F. 

If  D2  +  E2  —  &F  =  0,  the  only  real  values  satisfying  (6)  are 
x  =—  ^D,  y  —  —  \E  (footnote,  p.  52).  The  locus,  therefore,  is 
the  single  point  (—  £Z>,  —  J  E).  In  this  case  the  locus  of  (4) 
is  often  called  a  point-circle,  or  a  circle  whose  radius  is  zero. 

If  D2  +  E2  —  4  F  is  negative,  no  real  values  .satisfy  (6),  and 
hence  (4)  has  no  locus. 

115 


116 


ANALYTIC  GEOMETRY 


The  expression  D2  +  E2  —  4  F  is  called  the  discriminant  of  (4), 
and  is  denoted  by  ®.     The  result  is  given  by 

Theorem  I.    The  locus  of  the  equation 
(I)  v?  +  y2  +  Vx  +  Ey  +  F  =  0, 

whose  discriminant  is  ®  =  D2  +  Ez  ~  4  F,  is  determined  as  follows : 

(a)  When  ®  is  positive  the  locus  is  the  circle  whose  center  is 
(—%D,  —  %E)  and  whose  radius  is  r  =  £  Vl>2  +  E'2  —  4  F  =  %  V®. 

(b)  When  ®  1*5  zero  the  locus  is  the  point-circle  (—  £  Z),  —  ^  £). 

(c)  When  ®  ts  negative  there  is  no  locus. 

Corollary.    When  E  =  0  £/i<?  center  of  (I)  is  on  ^e  X-axis,  and 
when  D  =  0  £/ie  center  is  on  the  Y-axis. 

Whenever  in  what  follows  it  is  said  that  (I)  is  the  equation 
of  a  circle  it  is  assumed  that  ®  is  positive. 

Ex.  1.    Find  the  locus  of  the  equation  x2  +  y2  —  4x  +  8y  —  5  =  0. 

Solution.    The  given  equation  is  of 
the  form  (I),  where 

D  =  -  4,  E  =  8,  F  =  -  5, 
and  hence 


f*, 


\ 


\ 


A' 


The  locus  is  therefore  a  circle  whose 
center  is  the  point  (2,  —  4)  and  whose 
radius  is  \  VlOO  =  5. 

The  equation  Ax2  +  Bxy  +  Cy2 
-f  Dx  -f  Ey  +  F  =  0  is  called  the 
general  equation  of  the  second  degree 
in  x  and  y  because  it  contains  all 
possible  terms  in  x  and  y  of  the  second  and  lower  degrees.  This 
equation  can  be  reduced  to  the  form  (I)  when  and  only  when 
A  =  C  and  5  =  0.  Hence  the  locus  of  an  equation  of  the  second 
degree  is  a  circle  only  when  the  coefficients  of  x2  and  y2  are  equal 
and  the  xy-term  is  lacking. 

53.  Circles  determined  by  three  conditions.    The  equation  of 
any  circle  may  be  written  in  either  one  of  the  forms 


or 


THE  CIRCLE 


117 


Each  of  these  equations  contains  three  arbitrary  constants. 
To  determine  these  constants  three  equations  are  necessary,  and 
as  any  equation  between  the  constants  means  that  the  circle  sat- 
isfies some  geometrical  condition,  it  follows  that  a  circle  may  be 
determined  to  satisfy  three  conditions. 

Rule  to  determine  the  equation  of  a  circle  satisfying  three 
conditions. 

First  step.    Let  the  required  equation  be 

or 

as  m,ay  be  more  convenient. 

Second  step.  Find  three  equations  between  the  constants  a,  ft 
and  r  [or  D,  E,  and  F~\  which  express  that  the  circle  (1)  [or  (2)] 
satisfies  the  three  given  conditions. 

Third  step.  Solve  the  equations  found  in  the  second  step  for  a,  ft 
and  r  \_or  D,  E,  and  F]. 

Fourth  step.  Substitute  the  results  of  the  third  step  in  (1)  [or 
(2)].  The  result  is  the  required  equation. 

Ex.  1.  Find  the  equation  of  the  circle  passing  through  the  three  points 
Pi(0,  1),  P2(0,  6),  and  P8  (3,  0). 

Solution.    First  step.    Let  the  required  equa- 
tion be 


-3 


(3)  x2  +  y2  +  Dx  +  Ey  +  F  =  0. 

Second  step.  Since  PI,  P2,  and  P3  lie  on  (3), 
their  coordinates  must  satisfy  (3).  Hence  we 
have 

(4)  1  +  .E +  **  =  .(), 

(5)  36  +  6.E  +  F  =  0, 
and 

(6)  9  +  3.0+^  =  0. 

Third  step.    Solving  (4) ,  (5) ,  and  (6) ,  we  obtain 

E  =  -  7,  F  =  6,  D  =  -  5. 
Fourth  step.    Substituting  in  (3),  the  required  equation  is 


N 


118 


ANALYTIC  GEOMETRY 


By  Theorem  I  we  find  that  the  radius  is  f  V2*  and  the  center  is  the 
point  (|,  |). 

Ex.  2.    Find   the    equation   of   the    circle   passing   through   the   points 
PI  (0,  —  3)  and  P2  (4,  0)  which  has  its  center  on  the  line  x  +  2  y  =  0. 

Solution.    First  step.   Let  the  required  equation  be 
(7)  x2  -1-  y2  +  Dx  +  Ey  +  F  =  0. 

Second  step.    Since  PI  and  P2  lie  on  the  locus  of  (7),  we  have 

(8)  9-3^  +  ^  =  0 
and 

(9)  16 


Y\ 

/ 

"N, 

\ 

7 

V 

A 

'( 

0 

•^. 

~^ 

a 

\ 

5' 

fo) 

~^ 

^y 

\ 

(0, 

3) 

/ 

"^ 

PI 

•M^ 

lrl 

The  center  of  (7)  is     ---  ,  -  -,  and  since  it 


lies  on  the  given  line, 

-M-fH 


or 

(10)  D  +  2E  =  0. 

Third  step.    Solving  (8),  (9),  and  (10),  we  obtain 

D  =  -^-,E  =  |,  and  F  =  -  -™. 
Fourth  step.    Substituting  in  (7),  we  obtain  the  required  equation, 

x2  +  y2  -  Y35  +  $y-  ¥  =  °> 

or  5  x2  +  5  y2  -  14  x  +  7  y  —  24  =  0. 

The  center  is  the  point  (|,  —  T^),  and  the  radius  is  \  V29. 


PROBLEMS 


1  .  Find  the  equation  of  the  circle  whose  center  is 


(a)  (0,  1)  .a-nd  whose  radius  is  3. 

(b)  (-  2,  0)  and  whose  radius  is  2. 

(c)  (-  3,  4)  and  whose  radius  is  6. 
(e)  (<ar,  0)  and  whose  radius  is  a. 
(f  )  (0,  £)  and  whose  radius  is  /3. 
(g)  (°>  —  J8)  and  whose  radius  is  /3. 

*  The  radius  is  easily  obtained,  since  V2  is  the  length  of  the  diagonal  of  a  square 
whose  side  is  one  unit.  We  may  construct  a  line  whose  length  is  Vw  by  describing  a 
semicircle  on  a  line  whose  length  is  n  +  1  and  erecting  a  perpendicular  to  the  diameter 
one  unit  from  the  end.  The  length  of  that  perpendicular  will  be  V». 


Ans.  x2  +  yz  —  2  y  —  8  =  0. 

-4ns.  x2  +  y'2  +  4  x  =  0. 

Ans.  x2  +  y'2  +  6  x  -  8  y  =  0. 

Ans.  x2  +  y2  —  2  <xx  =  Q. 

^l?is.  x2  +  y2  —  2  j8y  =  0. 

^.ns.  x2  +  y2  +  2  /3y  =  0. 


THE  CIRCLE  119 


2.  Find  the  locus  of  the  following  equations. 

(a)  x2  +  2/2  _  Gx  _  16  =  0.  (f)  x2  +  2/2  _  QX  +  ±y  _  5  _  o. 

(b)  3x2  +  3y2  -  10 x  -  24  y  =  0.       (g)  (x  +  I)2  +  (y  -  2)2  =  0. 

(c)  x2  +  y2  =  0.  (h)  7x2  +  7  y2  -  4x  -  y  =  3. 

(d)  x2  +  y2  -  8x  -  Qy  +  25  =  0.       (i)  x2  +  y2  +  2  ox  +  2  fy  +  a2  +  ft2  =  0. 

(e)  x2  +  y2  -  2x  +  2 y  +  5  =  0.         (j)  x2  +  y'  +  IGx  +  100  =  0. 

3.  Find  the  equation  of  the  circle  which 

(a)  has  the  center  (2,  3)  and  passes  through  (3,  -  2). 

Ana.    x2  +  y2  -  4  x  -  6  y  -  13  =  0. 
;-$>)  passes  through  the  points  (0,  0),  (8,  0),  (0,  -  6)., 

Ana.    x2  +  £/2_8x  +  6y  =  0. 

(c)  passes  through  the  points  (4,  0),  (—  2,  5),  (0,  -  3). 

Ans.    19x2+ 192/2  + 2x- 47  y- 312  =  0. 

(d)  passes  through  the  points  (3,  5)  and  (—3,  7)  and  has  its  center  on 
the  JT-axis.  Ans.    x2  +  y2  +  4  x  -  46  =  0. 

(e)  passes  through  the  points  (4,  2)  and  (—6,  —  2)  and  has  its  center  on 
the  T-axis.  Ans,   x2  +  2/2  +  5  y  -  30  =  0. 

(f)  passes  through  the  points  (5,  —  3)  and  (0,  6)  and  has  its  center  on 
the  line  2x-3y-6  =  0.  Ans.    3  x2  +  3  y2  -  114  x  -  64  y  +  276  =  0. 

(g)  has  the  center  (—1,  —  5)  and  is  tangent  to  the  X-axis. 

Ans.  x2  +  2/2  _|_  2  x  +  10  y  +  1  =  0. 

(h)  passes  through  (1,  0)  and  (5,  0)  and  is  tangent  to  the  F-axis. 

Ana.    x2  +  ?/2-6x;t2V5y +  5  =  0. 
(i)  passes  through  (0,  1),  (5,  1),  (2,  -  3). 

Ans.    2x2  + 2?/2-10x  +  y-3  =  0. 
(j)  has  the  line  joining  (3,  2)  and  (—7,  4)  as  a  diameter. 

Ans.    x2  +  y2  +  4  x  -  6  y  —  13  =  0. 
(k)  has  the  line  joining  (3,  -  4)  and  (2,  —  5)  as  a  diameter. 

Ans.    x2  +  y2  _  5  x  +  9  y  +  26  =  0. 

(1)  which  circumscribes  the  triangle  formed  by  x  —  6  =  0,  x  +  2?/  =  0, 
and  x  -  2  y  =  8.  Ans.   2  x2  +  2  y2  _  21  x  +  8  y  +  60  =  0. 

(m)  passes  through  the  points  (1,  —  2),  (-  2,  4),  (3,  -  6).     Interpret  the 
result  by  the  Corollary,  p.  89. 

(n)  is  inscribed  in  the  triangle  formed  by4x  +  3y  —  12  =  0,  y  —  2  =  0, 
£-10  =  0.  Ana.    36x2  +  36y2  -  516x  +  QQy  +  1585  =  0. 

4.  Plot  the  locus  of  x2  +  y2  -  2x  +  4  y  +  k  =  0  for  k  =  0,  2,  4,  5  -  2,  -  4, 
—  8.     What  values  of  k  must  be  excluded ?  Ans.   k  >  5. 


120  ANALYTIC  GEOMETRY 

5.  What  is  the  locus  of  x2  +  y2  +  Dx  +  Ey  +  F  —  0  if  D  and  E  are  fixed 
and  F  varies  ? 

6.  For  what  values  of  k  does  the  equation  x2  +  y2  —  4 jc  +  2  ky  +  10  =  0 
have  a  locus  ?  Ans.   k  >  -f  Vo  and  A:  <  -  V6. 

7.  For  what  values  of  fc  does  the  equation  x2  +  y2  +  kx  +  F=Q  have  a 
locus  when  (a)  F  is  positive ;  (b)  F  is  zero  ;  (cj  F  is  negative  ? 

^Ins.    (a)  fc  >  2  \/F  and  fc  <  -  2  V*1 ;  (b)  and  (c)  all  values  of  k. 

8.  Find  the  number  of  point-circles  represented  by  the   equation  in 
problem  7.  Ans.    (a)  two ;  (b)  one  ;  (c)  none. 

9.  Find  the  equation  of  the  circle  in  oblique  coordinates  if  w  is  the  angle 
between  the  axes  of  coordinates. 

Ans.    (x  -  a)2  +  (y  -  /3)2  +  2  (x  -  a)  (y  ~  /3)  cos  <*  =  r2. 

10.  Write  an  equation  representing  all  circles  with  ^  the  radius  5  whose 
centers  lie  on  the  X-axis ;  on  the  F-axis. 

11.  Find  the  number  of  values  of  k  for  which  the  locus  of 

(a)  x2  +  y2  +  4  kx  -  2  y  +  5  k  =  0, 

(b)  x*  +  y*  +  4'fce  -  2y  -  k  =  0, 

(c)  x2  +  y2  +  4  kx  -  2  y  +  4  fc  =  0 

is  a  point-circle.  ^4ns.    (a)  two;  (b)  none  ;  (c)  one. 

12.  Plot   the   circles    z2  -f  y2  +  4  x  -  9  =  0,    z2  +  y2  -  4  x  -  9  =  0,    and 
xz  +  y°  +  4x  -9  +  k(x*  +  y*-4x-  9)  =  0  for  fc=±l,    ±3,    ±  1,    -5, 
—  £.     Must  any  values  of  k  be  excluded  ? 

13.  Plot  the  circles  x2  +  y2  +  4  x  =  0,  x2  +  y2  -  4  x  =  0,  and  x2  +  y2  +  4  « 
+  fc(z2  +  y2  —  4x)  =  0  for  the  values  of  k  in  problem  12.     Must  any  values 
of  k  be  excluded  ? 

14.  Plot   the    circles   x2  +  y2  +  4  x  +  9  =  0,    x2  +  y2  -  4  x  +  9  =  0,    and 
x2  +  ?/2  +  4x  +  9  +  fc(z2  +  y2  -  4x  +  9)  =  0    for   k  =  -  3,    -  |,    -  5,    -  |, 
~~  1»   ~~  f '   ~  1<     What  values  of  A;  must  be  excluded? 

54.  Systems  of  circles.    An  equation  of  the  form 

x2  +  if  +  Zta  +  £y  +  F  =  0 

will  define  a  system  of  circles  if  one  or  more  of  the  coefficients 
contain  an  arbitrary  constant.     Thus  the  equation 
ic2  +  ^2  _  r2  =  o 

represents  the  system  of  concentric  circles  whose  centers  are  at 
the  origin.  Very  interesting  systems  of  circles,  and  the  only 
systems  we  shall  consider,  are  represented  by  equations  analogous 
to  (XIII),  p.  110. 


THE  CIRCLE  121 

theorem  II.    Given  two  circles, 
Cl:x*  +  y*  +  D^x  +  E1y  +  Fl  =  Q 
Cz:x*  +  if  +  D2x  +  EJJ  +  F2  =  Oj 
\  the  locus  of  the  equation 
(II)  x2  +  y*  +  D&  +  EM  +  J\ 


is  a  circle  except  when  k  =  —  l.     In  this  case  the  locus  is  a  straight 
line. 

Proof.    Clearing  the  parenthesis  in  (II)   and   collecting  like 
terms  in  x  and  y,  we  obtain 


Dividing  by  1  -4-  k  we  have 


The  locus  of  this  equation  is  a  circle  (Theorem  I,  p.  116).  If, 
however,  k  =  —  1,  we  cannot  divide  by  1  +  k.  But  in  this  case 
equation  (II)  becomes 

(A  -  D2)aj  +  (El  -E2)y+  (F,  -  F»)  =  0, 

which  is  of  the  first  degree  in  x  and  y.     Its  locus  is.  then  a 
straight  line  called  the  radical  axis  of  C\  and  C2.  Q.E.D. 

Corollary  I.  The  center  of  the  circle  (II)  lies  upon  the  line 
joining  the  centers  of  Cx  and  C2  and  divides  that  line  into  seg- 
ments whose  ratio  is  equal  to  k. 

For  by  Theorem  I  (p.  116)  the  center  of  C\  is  PI  (  -    ~t  -     ^  )  and  of  Cz  ia 

-  \  *J      '  -j     / 

Pz(-    —  »  --  -  )  •    The  point  dividing  P\P>2  into  segments  whose  ratio  equals  Tc 

r-t  +*(-^)  -f  +K- 

is  (Theorem  VII,  p.  32)  the  point      -  ,          •  -  —  - 

^—  1  ~r"  ft^  1  "T~  A/ 

simplifying,  (  -  ff       ~  f2)'  wllteh  iS  th°  CeDter  °f  (H)' 


122 


ANALYTIC  GEOMETRY 


Corollary  II.    The  equation  of  the  radical  axis  of  C±  and  C2  is 

(D,  -D2)x  +  (E,  -Ez)y  +  (F,  -  F2)  =  0. 

Corollary  III.  The  radical  axis  of  two  circles  is  perpendicular  to 
the  line  joining  their  centers. 

Hint.  Find  the  line  joining  the  centers  of  Ct  and  C2  (Theorem  VII,  p.  88)  and  show 
that  it  is  perpendicular  to  the  radical  axis  hy  Corollary  III,  p.  78. 

The  system  (II)  may  have  three  distinct  forms,  as  illustrated 
in  the  following  examples.  These  three  forms  correspond  to  the 
relative  positions  of  Ci  and  C2,  which  may  intersect  in  two  points, 
be  tangent  to  each  other,  or  not  meet  at  all. 

Ex.  1.    Plot  the  system  of  circles  represented  by 

x2  +  y2  +  8x  -  9  +  k(x*  +  y2  -  4x  -  9)  =  0. 


Solution.    The  figure  shows  the  circles 

z2  +  y2-f8z-9=:0  and  z2  +  2/2-4z-9  =  0 
plotted  in  heavy  lines  and  the  circles  corresponding  to 
k  =  2,  5,  1,  i,  -  4,  -  f,  and  -  | ; 

these  circles  all  pass  through  the  intersection  of  the  first  two. 

The  radical  axis  of  the  two  circles  plotted  in  heavy  lines,  which  corre- 
sponds to  k  =  —  1,  is  the  Y-axis. 


THE  CIRCLE 


123 


Ex.  2.    Plot  the  system  of  circles  represented  by 

x2  +  y2  +  8x  +  k(x2  +  y2  -  4x)  =  0. 


X 


Solution.    The  figure  shows  the  circles 

X2  _j_  yi  +  8  x  =  0  and  x2  -f  y2  -  4  x  =  0 
plotted  in  heavy  lines  and  the  circles  corresponding  to 

k  =  2,  3,  |,  5,  1,  i,  -  7,  i,  -  4,  -  3,  and  -  \. 

These  circles  are  all  tangent  to  the  given  circles  at  their  point  of  tangency. 
The  locus  for  k  =  2  is  the  origin. 

Ex.  3.    Plot  the  system  of  circles  represented  by 

x2  +  y2  -  lOx  +  9  +  k  (x2  +  y2  -f  8x  +  9)  =  0. 


Solution.    The  figure  shows  the  circles 

x2  -f  y2  —  10  x  +  9  =  0  and  x2  +  y'2  +  8  x  +  9  =  0 
plotted  in  heavy  lines  and  the  circles  corresponding  to 
*  =  it  17,  i,  -  10,  -  TV,  and  -  -^. 


124  ANALYTIC  GEOMETRY 

These  circles  all  cut  the  dotted  circle  at  right  angles  (problem  7).     For  k=  |- 
the  locus  is  the  point-circle  (3,  0),  and  for  k  =  8  it  is  the  point-circle  (—  3,  0). 
In  all  three  examples  the  radical  axis,  for  which  k  =  —  1,  is  the  F-axis. 

PROBLEMS 

1.  If  Ci  and  C2  intersect  in  PI  and  P2,  the  system  (II)  consists  of  all 
circles  passing  through  PI  and  P2. 

2.  If  Ci  and  C2  are  tangent,  the  system  (II)  consists  of  all  circles  tan- 
gent to  Ci  and  <72  at  their  point  of  tangency. 

3.  The  radical  axis  of  two  intersecting  circles  is  their  common  chord, 
and  of  two  tangent  circles  is  the  common  tangent  at  their  point  of  tangency. 

4.  Find  the  equation  of  the  circle  passing  through  the  intersections  of 
the  circles  x2  +  y2  —  1  =  0  and  x2  +  y2  +  2  x  =  0  which  passes  through  the 
point  (3,  2).  Ans.    7  x2  +  7  y2  -  24  x  -  19  =  0. 

5.  Two  circles  x2  +  y2  +  DIX  +  E^y  +  FI  =  0  and  x2  +  y2  +  D2z  +  Ezy 
4>F2  =  0  intersect  at  right  angles  when  and  only  when  DiD2  +  EiE2  -  2  FI 
-  2  F2  =  0. 

Hint.  Construct  a  triangle  by  drawing  the  line  of  centers  and  the  radii  to  a  point  of 
intersection  I\. 

6.  The  equation  of  the  system  (II)  may  be  written  in  the  form  x2  +  y2 
+  k'x  +  F  =  0,  where  Fis  constant  and  k'  arbitrary,  if  the  axes  of  x  arid  y  be 
respectively  chosen  as  the  line  of  centers  and  the  radical  axis  of  Ci  and  C2. 

7.  The  system  in  problem  C  consists  of  all  circles  whose  intercepts  on  the 
F-axis  are  ±  V—  F  if  F<0,  which  are  tangent  to  the  F-axis  at  the  origin  if 
F  =  0,  and  which  intersect  the  circle  x2  +  y2  =  F  at  right  angles  if  P>0. 

8.  The  square  of  the  length  of  the  tangent  from  PI  (£1,2/1)  to  the  circle 
X2  +  y2  +  DX  +  Ey  +  F  =  0  is  Xi2  +  y^  +  Dxl  +  Ey±  +  F. 

Hint.  Construct  a  right  triangle  by  joining  Pt  and  the  point  of  tangency  to  the  center. 

9.  The  locus  of  points  from  which  tangents  to  two  circles  are  equal  is 
their  radical  axis. 

10.  Find  the  radical  axes  of  the  circles  x-  +  y2  -  4x  =  0,  x2  +  y2  +  6x 
—  8  y  ~  0,  and  x2  +  y'2  +  G  x  -  8  =  0  taken  by  pairs,  and  show  that  they 
meet  in  a  point. 

•11.  Show  that  the  radical  axes  of  any  three  circles  taken  by  pairs  meet 
in' a  point. 

12.  By  means  of  problem  11  show  that  a  circle  may  be  drawn  cutting  any 
three  circles  at  right  angles. 

13.  Show  that  the  radical  axis  of  any  pair  of  circles  in  the  system  (II) 
is  the  same  as  the  radical  axis  of  Oi  and  C2. 

14.  How  may  problem  11  be  stated  if  the  three  circles  are  point-circles  ? 


CHAPTER  VI 
POLAR   COORDINATES 


55.  Polar  coordinates.  In  this  chapter  we  shall  consider  a 
second  method  of  determining  points  of  the  plane  by  pairs  of 
real  numbers.  We  suppose  given  a  fixed  point  0,  called  the 
pole,  and  a  fixed  line  OA,  passing  through 
0,  called  the  polar  axis.  Then  any  point 
P  determines  a  length  OP  =  p  and  an 
angle  A  OP  =  0.  The  numbers  p  and  0 
are  called  the  polar  coordinates  of  P.  p  is 
called  the  radius  vector  and  0  the  vectorial 
angle.  The  yectorial  angle  0  is  positive 
or  negative  as  in  Trigonometry  (p.  11). 
The  radius  vector  is  positive  if  P  lies  on  the  terminal  line  of  0, 
and  negative  if  P  lies  on  that  line  produced  through  the  pole  O. 
Thus  in  the  figure  the  radius  vector  of  P  is  positive,  and  that  of  P'  is  negative. 

It  is  evident  that  every 
pair  of  real  numbers  (p,  0) 
determines  a  single  point, 
which  may  be  plotted  by 
the 

Rule  for  plotting  a  point 
whose  polar  coordinates 
(p)  0)  are  given. 

First  step.  Construct  the 
terminal  line  of  the  vecto- 
rial angle  0,  as  in  Trigo- 
nometry. 

Second  step.  If  the  radius 
vector  is  positive,  lay  off  a 
length  OP  =  p  on  the  terminal  line  of  0;  if  negative,  produce  the 

125 


126  ANALYTIC  GEOMETRY 

terminal  line  through  the  pole  and  lay  off  OP  eqi,al  to  the  numer- 
ical value  of  p.     Then  P  is  the  required  point. 

In  the  figure  on  p.  125  are  plotted  the  points  whose  polar  coordinates  are 


Every  point  P  determines  an  infinite  number  of  pairs  of  numbers  (p,  ff).  ___ 

The  values  of  0  will  differ  by  some  mul- 
tiple of  X,  so  that  if  0  is  one  value_of  6  the 
others  will  be  of  the  form  0  +  k  7t,  where  k  is 
a  positive  or  negative  integer.  The  values 
of  p  will  be  the  same  numerically,  but  will 
be  positive  or  negative,  if  P  lies  on  OB, 
according  as  the  value  of  6  is  chosen  so  that 
OB  or  OC  is  the  terminal  line.  Thus,  if 
OB  =  p  the  coordinates  of  B  may  be  written 
in  any  one  of  the  forms  (p,  0),  (—  p,  it  +  <£), 
(p, 


Unless  the  contrary  is  stated,  we  shall  always  suppose  that  0  is 
positive,  or  zero,  and  less  than  2  TT  ;  that  is,  0  <  6  <  2  TT. 

PROBLEMS 

'•"points  (4,  |),(0,  ^),(_,,'£),(4I  !),(-<,  IE)'. 

(o,  it). 

2.  Plot   the   points  (G,   ±£),  (-2,   ±f),  (3,  a),  (-4,  TT),  (6,  0), 
(-  6,  0). 

3.  Show  that  the  points  (/»,  ff)  and  (p,  —  6)  are  symmetrical  with  respect 
to  the  polar  axis. 

4.  Show  that  the  points  (p,  6),  (-  p,  6)  are  symmetrical  with  respect  to 
the  pole. 

5.  Show  that  the  points  (—  p,  it  —  ff)  and  (/>,  -0)  are  symmetrical  with  respect 
to  the  polar  axis. 

56.  Locus  of  an  equation.  If  we  are  given  an  equation  in  the 
variables  p  and  6,  'then  the  locus  of  the  equation  (p.  52)  is  a  curve 
such  that  : 

1.  Every  point  whose  coordinates  (p,  0)  satisfy  the  equation  lies 
on  the  curve. 

2.  The  coordinates  of   every  point  on  the  curve  satisfy  the 
equation. 


POLAR  COORDINATES 


127 


The  curve  may  be  plotted  by  solving  the  equation  for  p  and 
finding  the  values  of  p  for  particular  values  of  0  until  the  coor- 
dinates of  enough  points  are  obtained  to  determine  the  form  of 
the  curve. 

The  plotting  is  facilitated  by  the  use  of  polar  coordinate  paper,  which  enables 
ws  to  plot  values  of  6  by  lines  drawn  through  the  pole  and  values  of  p  by  circles 
having  the  pole  as  center.     The  tables  on  p.  14  are  to  be  used  in  constructing  . 
tables  of  values  of  p  and  0. 

In  discussing  the  locus  of  an  equation  the  following  points 
should  be  noticed. 

1.  The   intercepts  on  the  polar  axis  are  obtained  by   setting 
6  —  0  and  0  =  TT  and  solving  for  p. 

But  other  values  of  0  may  make  p  =  0  and  hence  give  a  point  on  the  polar  axis, 
namely,  the  pole. 

2.  The  curve  is  symmetrical  with  respect  to  the  pole  if,  when 
—  p  is  substituted  for  p,  only  the  form  of  the  equation  is  changed,^ 

3.  The  curve  is  symmetrical  with  respect  to  the  polar  axiVif, 
when  —  0  is  substituted  for  0,  only  the  form  of  the  equation  is 
changed. 

4.  The  directions  from  the  pole  in  which  the  curve  recedes  to 
infinity,   if   any,    are   found   by 

obtaining  those  values  of  0  for 
which  p  becomes  infinite. 

5.  The  method  of  finding  the 
values  of  0    which    must    be    ex- 
cluded, if   any,  depends   on  the 
given  equation. 

Ex.  1.  Discuss  and  plot  the  locus 
of  the  equation  p  =  10  cos  6. 

Solution.  The  discussion  enables  us 
to  simplify  the  plotting  and  is  there- 
fore put  first. 

1.  For  0  =  0  p  =  10,  and  for  0  =  it 
p  =  —  10.     Hence  the  curve  crosses  the 
polar  axis  10  units  to  the  right  of  the 
pole. 

2.  The    curve    is    symmetrical    with 
cos(-0)  =  cos0  (4,  p.  12). 


•       9 

p 

0 

P 

0 

10 

Tt 

0 

It 

I 

12 

9.7 

lit 

-  2.0 

12 

* 

-  6 

8.Z 

2* 
3 

-   5 

It 

Sit 

_   7 

I 

7 

4 

Tt 

5 

bit 
0 

-  8.7 

llTT 

-  9.7 

bit 

12 

12 

2.6 

Tt 

-10 

respect    to    the    polar    axis,    for 


128 


ANALYTIC  GEOMETRY 


3.  As  cos  0  is  never  infinite, 
the  curve  does  not  recede  to 
infinity.     Hence  the  curve  is  a 
closed  curve. 

4.  No  values  of  0  make  p 
imaginary. 

Computing  a  table  of  values 
we  obtain  the  table  on  p.  127. 

As  the  curve  is  symmetrical 
with  respect  to  the  polar  axis, 
the  rest  of  the  curve  may  be 
easily  constructed  without  com- 
puting the  table  farther ;  but 
as  the  curve  we  have  already 
constructed  is  symmetrical 
with  respect  to  the  polar  axis, 
no  new  points  are  obtained.  The  locus  is  a  circle. 

Ex.  2.    Discuss  and  plot  the  locus  of  the  equation  p2  =  a2  cos  2  0. 
Solution.  The  discussion  gives 
us  the  following  properties. 

1.  For  0  =  0or7rp  =  ±a. 
Hence  the  curve  crosses  the 
polar  axis  a  units  to  the  right 
and  left  of  the  pole. 

2.  The  curve    is   symmet- 
rical with  respect  to  the  pole. 

3.  It    is   also   symmetrical 
with    respect    to    the    polar 
axis,  for   cos  (—20)  =  cos  2  0 
(4,  p.  12). 

4.  p  does  not  become  infinite. 

5.  p    is    imaginary    when 
cos  2  0    is    negative.      cos  2  0 
is   negative    when    2  0   is    in 


e 

p 

9 

p 

0 

ifc 

Tt 

+.7a 

6 

Tt 

12 

dh  -93  a 

Tt 

4 

0 

the  second  or  third  quadrant;  that  is,  when 

STC  7t        7  it  bit 

— >20>-  or  — >20> 

222  2 

Hence  we  must  exclude  values  of  0  such  that 

3x  it       .  7  it  5rt 

_>0>-and— >0>—-. 
444  4 

The  accompanying  table  of  values  is  all  that 


need  be  computed  when  we  take  account  of  2,  3,  and  5. 


POLAR  COORDINATES 


129 


The  complete  -curve  is  obtained  by  plotting  these  points  and  the  points 
symmetrical  to  them  with  respect  to  the  polar  axis.  The  curve  is  called  a 
lemniscate.  In  the  figure  a  is  taken  equal  to  9.5. 

Ex.  3.    Discuss  and  plot  the  locus  of  the  equation 

2 

P  ~  1  +  cos  ^ 

Solution.  1.  For  6  =  0  p  =  1,  and  for  6 '=  it  p  =  oo  ;  so  the  curve  crosses 
the  polar  axis  one  unit  to  the  right  of  the  pole. 

2.  The  curve  is  not  symmetrical  with  respect  to  the  pole.     How  may  this 
be  inferred  from  1  ? 

3.  The  curve  is    symmetrical    with    respect   to   the   polar   axis,   since 
cos(-0)  =  cos0  (4,  p.  12). 

4.  p  becomes    infinite  when    1  +  cos  6  —  0    or    cos  6  =  -  1    and    hence 
6  =  7t.     The  curve  recedes  to  infinity  in  but  one  direction. 

5."^  is  never  imaginary. 

On  account  of  3  the  table  of  values  is  computed  only  to  6  =  7t,  and  the 
rest  of  the  curve  is  obtained  from  the  symmetry  with  respect  to  the  polar 

axis.     The  locus  is  a  parabola. 


e 

p 

e 

p 

0 

1 

lit 

2.7 

it 

1.02 

12 

12 

it 

1.07 

'~3~ 

4 

"6 

it 
I 

1.2 

^4~ 

6.7 

it 
3 

1.3 

57T 

6 

14 

5tf 
12 

1.6 

\\it 

50 

12 

it 

2 

2 

it 

CO 

5jr 


PROBLEMS 

Discuss  and  plot  the  loci  of  the  following  equations. 
1.  p  =  10.     0  =  tan~1l.  5.  p  sin  0  =  4. 

2p  "•  *-  »  ••  * 

.  p  ^^  o.     v  —  —  * 

3.  p  =  16  cos  0. 

4.  p  cos  0  =  6. 


130 


ANALYTIC  GEOMETRY 


8 


1  -  2  cos  0 
9.  p  =  a  sin  0. 

10.  p  =  a  (I  —  cose). 

11.  /3a sin  20  =  16. 

12.  p2  =  16sin20. 

13.  p2cos220  =  a2. 

14.  p  =  a  sin  2  0.     p  =  a  cos  2  0. 

8 


16.  p  cos  e  =  a  sin2  0. 

17.  />cos0  =  a  cos  20. 

18.  />  =  a  (4  -f  6  cos  0) 

for  6  =  3,  4,  6. 


15.  p  = 


1  —  e  cos  0 

for  e  =  1,  2, 


1  +  tan  0 

20.  p  =  a  sec  0  ±  & 

for  a  >  6,  a  =  6,  a  <  6. 

21.  p  =  ad. 

22.  p  =  a  sin  30.     /j  =  a  cos  30. 


23.  Prove  that  the  locus  of  an  equation  is  symmetrical  with  respect  to 

6  =  —  if  the  results  of  substituting  — \-  0  and 0  give  equations  which 

2  22 

differ  only  in  form. 

24.  Apply  the  test  for  symmetry  in  problem  23  to  the  loci  of  4,  5,  10,  11, 
and  12. 

57.  Transformation  from  rectangular  to  polar  coordinates. 

Let  OX  and  OF  be  the  axes  of  a  rectangular  system  of  coordi- 
nates, and  let  O  be 
the  pole  and  OX  the 
polar  axis  of  a  sys- 
tem of  polar  coor- 
dinates. Let  (x,  y) 
and  (p,  ff)  be  respec- 
~£*  tively  the  rectan- 
gular and  polar  coor- 
dinates of  any  point 
P.  It  is  necessary 

to  distinguish  two  cases  according  as  p  is  positive  or  negative. 
When  p  is  positive  (Fig.  1)  we  have,  by  definition, 


(2) 


cos  0  =  —j  sin  0  =  — 
P  P 


whatever  quadrant  P  is  in. 
Hence 


x  =  p  cos  0,  y  —  p  sin  9. 


POLAR  COORDINATES  131 

When  p  is  negative  (Fig.  2)  we  consider  the  point  P'  symmet- 
rical to  P  with  respect  to  0,  whose  rectangular  and  polar  coordi- 
nates are  respectively  (—  x,  —  y)  and  (—  p,  0).  The  radius  vector 
of  P',  —  /o,  is  positive  since  p  is  negative,  and  we  can  therefore  use 
equations  (1).  Hence  for  P' 

—  x  =—  p  cos  0,  —  y  =  —  p  sin  0; 
and  hence  for  P 

x  =      p  cos  0}       y  —  p  sin  0, 
as  before. 

Hence  we  have 

Theorem  I.  If  the  pole  coincides  with  the  origin  and  the  polar 
axis  with  the  positive  X-axis,  then 

where  (x,  ?/)  are  the  rectangular  coordinates  and  (p,  &)  the  polar 
coordinates  of  any  point. 

Equations  I  are  called  the  equations  of  transformation  from  rec- 
tangular to  polar  coordinates.  They  express  the  rectangular 
coordinates  of  any  point  in  terms  of  the  polar  coordinates  of 
that  point  and  enable  us  to  find  the  equation  of  a  curve  in  polar 
coordinates  when  its  equation  in  rectangular  coordinates  is  known, 
and  vice  versa. 

From  the  figures  we  also  have 

*         *         2  -1- 


+  v*  ± 

These  equations  express  the  polar  coordinates  of  any  point  in  terms 
of  the  rectangular  coordinates.  They  are  not  as  convenient  for  use 
as  (I),  although  the  first  one  is  at  times  very  convenient. 

Ex.  1.    Find  the  equation  of  the  circle  x2  +  y2  =  25  in  polar  coordinates. 

Solution.  Substitute  the  values  of  x  and  y  given  by  (I).  This  gives 
p2  cos2  e  +  p2  sin2  0  =  25,  or  (by  3,  p.  12)  p2  =  25  ;  and  hence  p  =  ±  5,  which 
is  the  required  equation.  It  expresses  the  fact  that  the  point  (/>,  6)  is  five 
units  from  the  origin. 


132  ANALYTIC  GEOMETRY 

Ex.  2.  Find  the  equation  of  the  lemniscate  (Ex.  2,  p.  128)  p2  =  a2  cos  2  0 
in  rectangular  coordinates. 

Solution.    By  14,  p.  13,  we  have 

/>2  =  a2(cos20-sin20). 

Multiplying  by  p2,  p4  =  a2  (p2  cos2  6  -  p2  sin2 6). 

From  (2)  and  (I),      (z2  +  y2)2  =  a2(x2  -  y2).   Ans. 

58.  Applications. 

Theorem  II.  The  general  equation  of  the  straight  line  in  polar 
coordinates  is 

(II)  p  (A  cos  0  -f  B  sin  ff)  +  C  =  0, 
where  A,  J5,  awd  C  are  arbitrary  constants. 

Proof.  The  general  equation  of  the  line  in  rectangular  coordi- 
nates is  (Theorem  II,  p.  77) 

Ax  -f  By  +  C  =  0. 
By  substitution  from  (I)  we  obtain  (II).'  Q.E.D. 

When  -4  =  0  the  line  is  parallel  to  the  polar  axis,  when  B  =.  0  it  is  perpendicular 
to  the  polar  axis,  and  when  C  =  0  it  passes  through  the  pole. 

In  like  manner  we  obtain 

Theorem  III.  The  general  equation  of  the  circle  in  polar  coordi- 
nates is 

(III)  p2  +  P(D  cos  0  +  E  sin  ff)  +  F  =  0, 
where  D,  E,  and  F  are  arbitrary  constants. 

Corollary.    If  the  pole  is  on  the  circumference  and  the  polar  axis 
passes  through  the  center,  the  equation  is 
p  —  2  r  cos  B  =  0, 
where  r  is  the  radius  of  the  circle. 

For  if  the  center  lies  on  the  polar  axis,  or  X-axis,  E  =  0  (Corollary,  p.  116) ; 
and  if  the  circle  passes  through  the  pole,  or  origin,  F  =  0.  The  abscissa  of  the 
center  equals  the  radius,  and  hence  (Theorem  I,  p.  116)  —  |  D  =  r,  or  D  =  —  2r. 
Substituting  these  values  of  Z>,  E,  and  F  in  (III)  gives  p  —  2  r  cos  0  =  0. 

Theorem  IV.  The  length  I  of  the  line  joining  two  points  P1  (p1}  0X) 
and  P2  (/o2,  02)  is  given  by 

(IV)  e  =  plt  +  pS-aplpt«»(9l-Oj. 


POLAR  COORDINATES'  133 

Proof.    Let  the  rectangular  coordinates  of  Px  and  P2  be  respec- 
tively (a?!,  2/x)  and  (x2,  ?/2).     Then  by  Theorem  I,  p.  131, 
#!  =  p!  cos  61,  x2  =  p2  cos  02, 
y!  =  pi  sin  01}  yz  =  p2  sin  02. 
By  Theorem  IV,  p.  24, 

Z2  =  (*!  -  *2)2  +  (yi  -  2/2)2, 
and  hence       I'2  =  (px  cos  01  —  p2  cos  02)2  +  (pl  sin  ^  —  p2  sin  02)2. 

Removing  parentheses  and  using  3,  p.  12,  and  11,  p.  13,  we 
obtain  (IV).  Q.E.D. 

PROBLEMS 

1.  Transform  the  following  equations  into  polar  coordinates  and  plot 
their  loci. 

(a)  x  —  3  y  =  0.  Ans.   6  =  tan-1^. 

(b)  2/-f5  =  0.  Ans.    p  =  -m  —  • 

sin  Q 

(c)  x2  +  y'2  =  16.  Ans.   p  =  ±  4. 

(d)  x2  +  y2  —  ax  =  0.  -4ns.   p  =  a  cos  0. 


(f)  x2  —  y*  =  a2.  Ans.   /o2  cos  2  0  =  a2. 

(g)  x  cos  w  +  y  sin  w  —  p  =  0.  Ans.   p  cos  (0  -  w)  —  p  =  0. 


(h)  (1  -  e2)x2  +  ?y2  -  2  e2px  -  e2p2  =  0. 


1  —  e  cos  0 
(i)  2  xy  +  4  y2  -  8z  +  9  =  0.    Ans.  p2  (sin  20  +  4  sin20)  -  8p  cos  0  -f  9  =  0. 

2.  Transform  equations  1  to  21,  p.  129,  into  rectangular  coordinates. 

3.  Find  the  polar  coordinates  of  the  points  (3,  4),  (-  4,  3),  (5,  -  12), 
(4,  5). 

4.  Find  the  rectangular  coordinates  of  the  points  (  5,  —  J,  (  —  2,  --  J> 
(3,  7t). 

5.  Transform  into  rectangular  coordinates  p  =  -  —  -- 

1  -  e  cos  0 

59.  Equation  of  a  locus.  The  equation  of  a  locus  may  often 
be  found  with  more  ease  in  polar  than  in  rectangular  coordinates, 
especially  if  the  locus  is  described  by  the  end  of  a  line  of  variable 
length  revolving  about  a  fixed  point.  The  steps  in  the  process  of 
finding  the  polar  equation  of  a  locus  correspond  to  those  in  the 
Rule  on  p.  46. 


134 


ANALYTIC   GEOMETRY 


Ex.  1.    Find  the  locus  of  the  middle  points  of  the  chords  of  the  circle 
C :  p  —  2  r  cos  0  =  0  which  pass  through  the  pole  which  is  on  the  circle. 

Solution.   Let  P(p,  0)  be  any  point  on  the  locus.     Then,  by  hypothesis, 

OP  =  $  OQ, 
where  Q  is  a  point  on  C. 

But       OP  =  p  and  OQ  =  2  r  cos  0. 
Hence  p  =  r  cos  0. 

From  the  Corollary  (p.  132)  it  is  seen  that 
the  locus  is  a  circle  described  on  the  radius 
of  C  through  O  as  a  diameter. 

Ex.  2.    The  radius  of  a  circle  is  prolonged  a  distance  equal  to  the  ordinate 
of  its  extremity.     Find  the  locus  of  the  end  of  this  line. 

Solution.    Let  r  be  the  radius  of  the  circle,  let  its  center  be  the  pole,  and 
let  P  (p,  0)  be  any  point  on  the  locus.     Then, 
by  hypothesis, 


OP  =  OB  +  CB. 
But  OP  =  />, 

OB  =  r, 
and  CB  =  r  sin  0. 

Hence  the  equation  of  the  locus  of  P  is 

p  =  r  +  r  sin  6. 
The  locus  of  this  equation  is  called  a  cardioid. 


P(P,6) 


PROBLEMS 

1.  Chords  passing  through  a  fixed  point  on  a  circle  are  extended  their 
own  lengths.     Find  the  locus  of  their  extremities. 

Ans.    A  circle  whose  radius  is  a  diameter  of  the  given  circle. 

2.  Chords  of  the  circle  p  =  10  cos  6  which  pass  through  the  pole  are 
extended  10  units.     Find  the  locus  of  the  extremities  of  these  lines. 

Ans.   p  =  10  (1  -f  cos  0). 

3.  Chords  of  the  circle  p  =  2  a  cos  0  which  pass  through  the  pole  are 
extended  a  distance  26.    Find  the  locus  of  their  extremities. 

Ans.   p  =  2  (6  +  a  cos  0). 


POLAR  COORDINATES  135 

4.  Find  the  locus  of  the  middle  points  of  the  lines  drawn  from  a  fixed 
point  to  a  given  circle. 

Hint.   Take  the  fixed  point  for  the  pole  and  let  the  polar  axis  pass  through  the  center 
the  circle. 

Ans.    A  circle  whose  radius  is  half  that  of  the  given  circle  and  whose 
center  is  midway  between  the  pole  and  the  center  of  the  given  circle. 


" 


5.  A  line  is  drawn  from  a  fixed  point  O  meeting  a  fixed  line  in  PI.     Find 
the  locus  of  a  point  P  on  this  line  such  that  OPj  •  OP  =  a2.    Ans.    A  circle. 

6.  A  line  is  drawn  through  a  fixed  point  O  meeting  a  fixed  circle  in  Pj 
and  PZ.     Find  the  locus  of  a  point  P  on  this  line  such  that 

r\~p     o~p 
OP  =  2        l '       2  .       Ans.   A  straight  line. 


CHAPTER  VII 


TRANSFORMATION  OF  COORDINATES 

60.  When  we  are  at  liberty  to  choose  the  axes  as  we  please 
we  generally  choose  them  so  that  our  results  shall  have  the  sim- 
plest possible  form.     When  the  axes  are  given  it  is  important 
that  we  be  able  to  find  the  equation  of  a  given  curve  referred  to 
some  other  axes.     The  operation  of  changing  from  one  pair  of 
axes  to  a  second  pair  is  known  as  a  transformation  of  coordinates. 
We  regard  the  axes  as  moved  from  their  given  position  to  a  new 
position  and  we  seek  formulas  which  express  the  old  coordinates 
in  terms  of  the  new  coordinates. 

61.  Translation  of  the  axes.    If  the  axes  be  moved  from  a  first 
position  OX  and  OF  to  a  second  position  O'X'  and  O'Y'  such  that 
O'X'  and  O'Y'  are  respectively  parallel  to  OX  and  OF,  then  the 
axes  are  said  to  be  translated  from  the  first  to  the  second  position. 

Let  the  new  origin  be   0' (h,  k)   and  let  the   coordinates  of 

any  point  P  before  and  after  the 
translation  be  respectively  (#,  y) 
and  (aJf,  y').  Projecting  OP  and 
00'P,on  OX,  we  obtain  (Theorem 
XI,  pi  41) 

x  =  x'  +  h. 

Similarly,  y  =  y'  +  k. 

V  Hence, 

Theorem  I.  If  the  axes  be  translated  to  a  new  origin  (h,  7c),  and 
if  (x,  y)  and  (x',  y')  are  respectively  the  coordinates  of  any  point  P 
before  and  after  the  translation,  then 

=  K>  +h, 


136 


A7 


MX 


TRANSFORMATION  OF  COORDINATES 


137 


TO 


Equations  (I)  are  called  the  equations  for  translating  the  axes, 
find  the  equation  of  a  curve  referred  to  the  new  axes  when 
its  equation  referred  to  the  old  axes  is  given,  we  substitute  the 
values  of  x  and  y  given  by  (I)  in  the  given  equation.  For  the 
given  equation  expresses  the  fact  that  P  (x,  y)  lies  on  the  given 
curve,  and  since  equations  (I)  are  true  for  all  values  of  (x,  y),  the 
new  equation  gives  a  relation  between  x'  and  y1  which  expresses 
that  P(x',  y')  lies  on  the  curve  and  is  therefore  (p.  46)  the  equa- 
tion of  the  curve  in  the  new  coordinates. 

Ex.  1.    Transform  the  equation 


when  the  axes  are  translated  to  the  new  origin  (3,  —  2). 

Solution.    Here  h  =  3  and  k  =  —  2,  so  equations  (I)  become 
x  z  —  x'  +  3,  y  =  y'  —  2. 

Substituting  in  the  given  equation,  we 
obtain 


\ 


X' 


or,  reducing,    z/2  +  y'2  =  25. 

This  result  could  easily  be  foreseen. 
For  the  locus  of  the  given  equation  is 
(Theorem  I,  p.  116)  a  circle  whose  center  is 
(3,  —2)  and  whose  radius  is  5.  When  the 
origin  is  translated  to  the  center  the  equa- 
tion of  the  circle  must  necessarily  have 
the  form  obtained  (Corollary,  p.  51). 

PROBLEMS 

1.  Find  the  new  coordinates  of  the  points  (3,  —  5)  and  (—4,  2)  when  the 
axes  are  translated  to  the  new  origin  (3,  6). 

2.  Transform  the  following  equations  when  the  axes  are  translated  to  the 
new  origin  indicated  and  plot  both  pairs  of  axes  and  the  curve. 

(a)  3z-4y  =  6,  (2,  0).  Ans.    3z/-4y/  =  0. 

(c)  ?/2-6z  +  9  =  0,  (3,  0).  Ans.   y^  =  Qx'. 

(d)  z2  +  2/2_i  =  o,  (-3,  -2).  Ans. 

(e)  y2  —  2  kx  +  k2  =  0,  (  - ,  0  V  -4ns. 

(f)  z2-4y24-8z+24y-20=0,  (-4,3).   Ans. 


138 


ANALYTIC  GEOMETRY 


3.  Derive  equations  (I)  if  O'  is  in  (a)  the  second  quadrant ;  (b)  the  third 
quadrant ;  (c)  the  fourth  quadrant. 

62.  Rotation  of  the  axes.  Let  the  axes  OX  and  OF  be  rotated 
about  0  through  an  angle  B  to  the  positions  OX'  and  OF'.  The 
equations  giving  the  coordinates  of  any  point  referred  to  OX  and 
OF  in  terms  of  its  coordinates  referred  to  OX'  and  OF'  are  called 
the  equations  for  rotating  the  axes. 

Theorem  II.    The  equations  for  rotating  the  axes  through  an  anyle 

6  are 
-p  C^  =  acf  cos  9  —  y*  sin  0, 


[y  =  a?'  sin  0  +  y'  cos  9. 

Proof.    Let  P  be  any  point 
whose  old   and    new   coordi- 

^   nates  are  respectively  (x,  y) 

x     and   (x',  y1).     Draw  OP  and 
draw  PM '  perpendicular  to  OX'.     Project  OP  and  OM'P  on  OX. 

The  proj.  of  OP  on  OX  =  x.  (Theorem  III,  p.  24) 

The  proj.  of  OM'  on  OX  =  x'  cos  0.  (Theorem  II,  p.  23) 

/  \ 

The  proj.  of  M'P  on  OX  =  y'cos  (|  +  0  I    (Theorem  II,  p.  23) 

\  / 

=  -y'sin0.  (by  6,  p.  13) 

Hence  (Theorem  XI,  p.  41) 

x  =  x'  cos  6  —  y'  sin  0. 

In  like  manner,  projecting  OP  and  OM'P  on  OF,  we  obtain 

j  \ 

y  =  x'  cos  (  —   —  0  )  +  y'  cos  0 

\.  / 

=  x'  sin0  +  ?/'cos0. 


Q.E.D. 


If  the  equation  of  a  curve  in  #  and  y  is  given,  we  substitute 
from  (II)  in  order  to  find  the  equation  of  the  same  curve  referred 
to  OX'  and  OF'. 


TRANSFORMATION  OF  COORDINATES  139 

Ex.  1.    Transform  the  equation  x2  -  y2  =  16  when  the  axes  are  rotated 

$K 


, 
through  — 


Solution.    Since 


4      2 

7f  1 

and      cos  —  =  —  -» 

4      V2 

equations  (II)  become 


- 

V5 


Substituting    in    the    given 
equation,  we  obtain 


X 


or,  simplifying, 


x'y'  +  8  =  0. 


PROBLEMS 


1.  Find  the  coordinates  of  the  points  (3,  1),  (—  2,  6),  and  (4,  —  1)  when 

the  axes  are  rotated  through  —  • 

2 

2.  Transform  the  following  equations  when  the  axes  are  rotated  through 
the  indicated  angle.     Plot  both  pairs  of  axes  and  the  curve. 

„    ft 


, 
4 


?/'  =  0. 

Ans.   x'2  =  4. 
Ans.   x'2  =  4  y'. 


(b)  x2  + 

(c)  y2  = 

(d)  x2  + 

(e)  x2  +  y2  =  r~,  6.  Ans. 

ff )  x2  4-  2  xy  +  y2  -f  4  x  —  4  y  =  0, .  Ans. 

4 

3.  Derive  equations  (II)  if  6  is  obtuse. 

63.  General  transformation  of  coordinates.  If  the  axes  are 
moved  in  any  manner,  they  may  be  brought  from  the  old  position 
to  the  new  position  by  translating  them  to  the  new  origin  and 
then  rotating  them  through  the  proper  angle. 


140 


ANALYTIC  GEOMETRY 


Theorem  III.  If  the  axes  be  translated  to  a  new  origin  (A,  k)  and 
then  rotated  through  an  angle  0,  the  equations  of  the  transforma- 
tion of  coordinates  are 

/j-rjv'  f  a?  =  a?'  cos  0  —  t/f  sin  0  +  h, 

\y  =  x1  sin  0  -f  y'  cos  0  -f  k. 

Proof.    To  translate  the  axes  to  O'X"  and  O'Y"  we  have,  by  (I), 

x  =  #*  -f  7*, 

y  =  y*  +  &> 

where  (#*,  ?/')  are  the  coordi- 
nates of  any  point  P  referred 
to  O'X*  and  O'Y*. 

To  rotate  the  axes  we  set, 

by  (ii), 

x*  =  je"cos  0  —  7/f/sin  0, 
yt  =  a  "sin  0  +  y'cos  0. 


Substituting  these  values  of  x*  and  ?/*,  we  obtain  (III). 


Q.E.D. 


64.  Classification  of  loci.  The  loci  of  algebraic  equations 
(p.  10)  are  classified  according  to  the  degree  of  the  equations. 
This  classification  is  justified  by  the  following  theorem,  which 
shows  that  the  degree  of  the  equation  of  a  locus  is  the  same  no 
matter  how  the  axes  are  chosen. 

Theorem  IV.  The  degree  of  the  equation  of  a  locus  is  unchanged 
by  a  transformation  of  coordinates. 

Proof.  Since  equations  (III)  are  of  the  first  degree  in  x'  and 
?/',  the  degree  of  an  equation  cannot  be  raised  when  the  values  of 
x  and  y  given  by  (III)  are  substituted.  Neither  can  the  degree 
be  lowered ;  for  then  the  degree  must  be  raised  if  we  transform 
back  to  the  old  axes,  and  we  have  seen  that  it  cannot  be  raised 
by  changing  the  axes.* 

As  the  degree  can  neither  be  raised  nor  lowered  by  a  trans- 
formation of  coordinates,  it  must  remain  unchanged.  Q.E.D. 


*  This  also  follows  from  the  fact  that  when  equations  (III)  are  solved  for  $'  and  y'  the 
results  are  of  the  first  degree  in  x  and  y. 


TRANSFORMATION  OF  COORDINATES  141 


65.  Simplification  of  equations  by  transformation  of  coordi- 
nates. The  principal  use  made  of  transformation  of  coordinates 
to  discuss  the  various  forms  in  which  the  equation  of  a  curve 
y  be  put.  In  particular,  they  enable  us  to  deduce  simple  forms 
which  an  equation  may  be  reduced. 

Rule  to  simplify  the  form  of  an  equation. 

first  step.  Substitute  the  values  of  x  and  y  given  by  (I)  [or  (II)  J 
and  collect  like  powers  of  x'  and  y'. 

Second  step.  Set  equal  to  zero  the  coefficients  of  two  terms 
obtained  in  the  first  step  which  contain  h  and  k  (or  one  coeffi- 
cient containing  ff). 

Third  step.  Solve  the  equations  obtained  in  the  second  step  for 
h  and  k*  (or  0). 

Fourth  step.  Substitute  these  values  for  h  and  k  (or  0)  in  the 
result  of  the  first  step.  The  result  will  be  the  required  equation. 

In  many  examples  it  is  necessary  to  apply  the  rule  twice  in 
order  to  rotate  the  axes,  and  then  translate  them,  or  vice  versa. 
It  is  usually  simpler  to  do  this  than  to  employ  equations  (III) 
in  the  Rule  and  do  both  together.  Just  what  coefficients  are 
set  equal  to  zero  in  the  second  step  will  depend  on  the  object 
in  view. 

It  is  often  convenient  to  drop  the  primes  in  the  new  equation 
and  remember  that  the  equation  is  referred  to  the  new  axes. 

Ex.  1.  Simplify  the  equation  y2  —  8x  +  6  y  +  .17  =  0  by  translating  the 
axes. 


Solution.    First  step.    Set  x  =  x'  +  h  and  y  =  y'  +  k. 
This  gives  (yf  +  fc)»  -  8  (a/  +  h)  +  6  (yf  +  k)  +  17  =  0,  or 


(1) 


6 


+  A:2 
-Sh 
+  6* 

+  17 


t  =  0. 


*  It  may  not  be  possible  to  solve  these  equations  (Theorem  IV,  p.  81). 

t  These  vortical  bars  play  the  part  of  parentheses.  Thus  2  k  +  6  is  the  coefficient  of  y' 
and  !;•  —  8/*  +  Gk  +  17  is  the  constant  term.  Their  use  enables  us  to  collect  like  powers 
of  x'  and  y'  at  the  same  time  that  we  remove  the  parentheses  in  the  preceding  equation. 


142 


ANALYTIC  GEOMETRY 


Second  step.  Setting  the  coefficient  of  y'  and  the  constant  term,  the  only 
coefficients  containing  h  and  k,  equal  to  zero,  we 
obtain 

(2)  2fc+6  =  0, 

Third  step.  Solving  (2)  and  (3)  for  h  and  Jk, 
we  find 

k  =  -3,  h  =  l. 

Fourth  step.  Substituting  in  (1),  remember- 
ing that  h  and  k  satisfy  (2)  and  (3),  we  have 

The  locus  is  the  parabola  plotted  in  the  figure 
which  shows  the  new  and  old  axes. 

Ex.  2.  Simplify  z2  +  4y2-2z  —  16  y  +  1  =  0 
by  translating  the  axes. 

Solution.  First  step.  Set  z  =  z'  -f  A,  y = y' + k. 
This  gives 


Y' 

^Y 

^f 

.X 

/ 

/ 

r 

/ 

/ 

0 

/ 

x 

/ 

O 

(1 

-3) 

x' 

\ 

\ 

\ 

\ 

\ 

s 

^ 

^ 

[ 

(4) 


-2 


-16 


A2 


=0. 


-2h 
-16k 
+  1 

Second  step.    Set  the  coefficients  of  x'  and  y'  equal  to  zero.     This  gives 

2/i-2  =  0,  8&-16  =  0. 
Third  step.    Solving,  we  obtain 

h  =  1,  A;  =  2. 
Fourth  step.    Substituting  in  (4)  ,  we  obtain 


Plotting  on  the  new  axes,  we  obtain  the 
figure. 

Ex.  3.    Remove  the  zy-term  from  x2  +  4  xy  +  y2  —  4  by  rotating  the  axes. 
Solution.    First  step.    Set  x  =  x'  cos  Q  —  y'  sin  6  and  y  =  x'  sin  0  -f  y"  cos  0, 
whence 


COS20 

+  4  sin  0  cos  0 
+  sin2  6 


—  2  sin  6  cos  0 

+  4  (cos2  0  -  sin2  0) 

+  2  sin  0  cos  0 


— 4sin0cos0 
-f  cos20 


7/2  =  4, 


or,  by  3,  p.  12,  and  14,  p.  13, 

(5)  (1  +  2sin20)a/2  +  4  Cos20-  x'y'  +'(1  -  2  sin  2  0)  y'2  =  4. 


TRANSFORMATION  OF  COORDINATES 


143 


Second  step.    Setting  the  coefficient  of  xfy'  equal  to  zero,  we  have 

cos20  =  0.  -*  '    Y 

Third  step.    Hence 


Fourth  step.  Substituting  in 
(5),  we  obtain,  since  sin  —  =  1 
(P-  14), 


The  locus  of  this  equation  is 
the  hyperbola  plotted  on  the 
new  axes  in  the  figure. 


/ 


5 


From  cos  2  0  =  0  we  get,  in  general,  2  6  =  — h  mt,  where  n  is  any  positive 

2 

or  negative  integer,  or  zero,  and  hence  6  =  — h  n  —  •     Then  the  xy-term  may 

be  removed  by  giving  6  any  one  of  these  values.     For  most  purposes  we 
choose  the  smallest  positive  value  of  0  as  in  this  example. 


Ex.  4.    Simplif 

/"  ic^  •"I"  6  iK^   1   i  o  /j*  .,    4-  ?/ 
&oZu£ion. 

+  4  =  0  by 
First  step. 

translating  the  axes. 
Set 

~h 

x  =  x'  +  h,  y  =  y/  +  Jc. 

We  obtain 

x'2+    3  A2 
+  12  h 

+  12 

sp.    Set  equ 
>  constant  t( 
3^  +  ( 

.    Solving, 

x'  —  4  ?/'  +       hs 
+    6  A2 
+  12  h 

+    4 
al  to  zero  the  coefl 
»rm.     This  gives 
,  =  0, 
-  4  k  +  4  =  0. 

^0, 

(6)  z'3  +  3/i 

1 

+  6 

5cient 

I 

/ 

/ 

r~ 

> 

o 

/ 

i)' 

HJrij 

i 

t 

P 

h8  4 

/ 

-            Third  step 

/ 

ff 

p.    Substituting  in  (6),  we  obtain 

Fourth  ste 

whose  locus  is  the  cubical  parabola  in  the  figure. 


144  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Simplify  the  following  equations  by  translating  the  axes.     Plot  both 
pairs  of  axes  and  the  curve. 

(a)  x2  +  6x  +  8  =  0.  Ans.  x'2  =  1. 

(b)  x2  -  4  y  +  8  =  0.  Ans.  x'*  =  4  y'. 

(c)  x2  +  ?/2  +  4  x  -  6  y  -  3  =  0.  ^ins.  x'2  -f  y'2  =  16. 

(d)  ?/2-6x-10y  +  19  =  0.  Ans.  y'2  =  6x'. 

(e)  x2  -  y2  +  8x  -  14 y  -  33  =  0.  Ana.  x'2  -  y/2  =  0. 

(f)  x2  +  4  y2  -  16  x  +  24  y  +  84  =  0.  4na.  x'2  +  4  y'2  =  16. 

(g)  y3  +  8  x  —  40  =  0.  Ans.  8  x'  +  ?/'3  =  0. 
(h)  x3  -  y2  +  14  y  -  49  =  0.  ^Ins.  ?//2  =  x/3. 

(i)  4x2-4xy4- y2-40x+ 20y+99  =  0.        Ana.    (2x'  -  y')*  -  1  =  0. 

2.  Remove  the  xy-term  from  the  following  equations  by  rotating  the  axes. 
Plot  both  pairs  of  axes  and  the  curve. 

(a)  x2  -  2 xy  +  y2  =  12.  Ans.  y/2  =  6. 

(b)  x*  -  2 xy  +  y2  +  8x  -f  8 y  =  0.  Ans.  V2 y'2  +  8x'  =  0. 

(c)  xy  =  18.  .4ns.  x'2  -  y'2  =  36. 

(d)  25 x2  +  14  xy  -f  25 y2  =  288.  Ans.  16  x/2  -f  9  y/2  =  144. 

(e)  3x2  -  lOxy  -f  3 y2  =  0.  Ana.  x/2  -  4 y/2  =  0. 

(f )  6  x2  +  20  V3  xy  +  26  y2  =  324.  ^Ins.  9  x/2  -  y/2  =  81. 

66.  Application  to  equations  of  the  first  and  second  degrees. 

In  this  section  we  shall  apply  the  Eule  of  the  preceding  section 
to  the  proof  of  some  general  theorems. 

Theorem  V.    By  moving  the  axes  the  general  equation  of  the  first 
degree, 

Ax  +  By  +  C  =  0, 

may  be  transformed  into  x'  =  0. 

Proof.    Apply  the  Eule  on  p.  141,  using  equations  (HI). 
Set  x  =  x'  cos  6  —  y1  sin  0  +  h, 

y  —  x'  sin  6  +  y1  cos  0  +  k. 
This  gives 


(1)  A  cos  e 

+  B  sin  0 


x'  —  A  sin  0 
+  B  cos  0 


+  Bk 

+  C 


=  0. 


TRANSFORMATION  OF  COORDINATES 


145 


Setting  the  coefficient  of  y'  and  the  constant  term  equal  to  zero 
gives 

(2)  —  A  sin  6  +  B  cos  0  =  0, 

(3)  Ah  +  Bk  +  C  =  0. 


From  (2), 


tan0  =  ->    or  0  =  tan-M- 


From  (3)  we  can  determine  many  pairs  of  values  of  h  and  k. 
One  pair  is 

h= ->      k  =  0. 


Substituting  in  (1)  the  last  two  terms  drop  out,  and  dividing 
by  the  coefficient  of  x'  we  have  left  x'  =  0.  Q.E.D. 

We  have  moved  the  origin  to  a  point  (h,  k)  on  the  given  line 
L,  since  (3)  is  the  condition  that  (h,  k)  lies  on  the  line,  and  then 
rotated  the  axes  until  the  new  axis  of 
y  coincides  with  L.  The  particular 
point  chosen  for  (h,  k)  was  the  point  0' 
where  L  cuts  the  A'-axis. 

This  theorem  is  evident  geometric- 
ally. For  x1  =  0  is  the  equation  of 
the  new  F-axis,  and  evidently  any  line 
may  be  chosen  as  the  F-axis.  But  the  theorem  may  be  used  to 
prove  that  the  locus  of  every  equation  of  the  first  degree  is  a 
straight  line,  if  we  prove  it  as  above,  for  it  is  evident  that  the 
locus  of  x'  =  0  is  a  straight  line. 

Theorem  VI.    The  term  in  xij  may  always  be  removed  from  an 
equation  of  the  second  degree, 


Ax*  +  Bxy  +  Cya  +  Dx  +  Ey  +  F  =  0, 
by  rotating  the  axes  through  an  angle  0  such  that 


(VI) 


tan20  = 


A-C 


146 


ANALYTIC  GEOMETRY 


Proof.    Set 
and 

This  gives 


sin  0  cos 

-f  C  sin2  0 


x  =  x'  cos  0  —  yf  sin  0 
y  =  x'  sin  0  -f  y'  cos  0. 


—  2J.  sin  0  cos  0 
-}-.B(cos20  —  sin20) 
+  2  C  sin  0  cos  £ 


—  B  sin  0  cos 
+  C  cos2  0 


4-  D  cos  0 
-f  E  sin  0 


x1  —  D  sin 
+  £  cos 


y'  +  .F  =  0. 


Setting  the  coefficient  of  x'y'  equal  to  zero,  we  have 

(C  —  A)  2  sin  0  cos  0  +  B  (cos2  0  —  sin2  0)  =  0, 
or  (14,  p.  13),  (C  -  A)  sin  2  0  +  B  cos  2  0  =  0. 

7? 


Hence 


tan  2  0  = 


—  C 


If  0  satisfies  this  relation,  on  substituting  in  (4)  we  obtain  an 
equation  without  the  term  in  xy.  Q.E.D. 

Corollary.  In  transforming  an  equation  of  the  second  degree  by 
rotating  the  axes  the  constant  term  is  unchanged  unless  the  new 
equation  is  multiplied  or  divided  by  some  constant. 

For  the  constant  term  in  (4)  is  the  same  as  that  of  the  given  equation. 

Theorem  VII.  The  terms  of  the  first  degree  may  be  removed  from 
an  equation  of  the  second  degree, 

Ax*  +  Bxy  +  C?/2  -f-  Dx  +  Ey  -f  F  =  0,     . 

by  translating  the  axeSj  provided  that  the  discriminant  of  the  terms 
of  the  second  degree,  A  =  B2  —  4  A  C,  is  not  zero. 

Proof.     Set  x  =  x'  +  h,  y  =  y'  +  k. 

This  gives 


(5)      Ax'2  -f  Bxy  -f 


+  2.4A 

+  £& 


x'  +  5/i, 
+  2C& 
-f  # 


y 


+  Bhk 

+  Dh 
+  Ek 


=  0. 


TRANSFORMATION  OF  COORDINATES  147 

Setting  equal  to  zero  the  coefficients  of  x'  and  y1,  we  obtain 
'(6)  2Ah  +  Bk  +  Z>  =  0, 

(7)  Bh  +  2  Ck  +  E  =  0. 

These  equations  can-  be  solved  for  h  and  k  unless  (Theorem 

IV,  p.  81) 

2A  _    B 

B    '"2C* 
or  B*  -  4  A  C  =  0. 

If  the  values  obtained  be  substituted  in  (5),  the  resulting  equa- 
tion will  not  contain  the  terms  of  the  first,  degree.  Q.E.D. 

Corollary  I.  If  an  equation  of  the  second  degree  be  transformed 
by  translating  the  axes,  the  coefficients  of  the  terms  of  the  second 
degree  are  imchanged  unless  the  new  equation  be  multiplied  or 
divided  by  some  constant. 

For  these  coefficients  in  (5)  are  the  same  as  in  the  given  equation. 

Corollary  II.  When  A  is  not  &ero  the  locus  of  an  equation  of  the 
second  degree  has  a  center  of  symmetry. 

For  if  the  terms  of  the  first  degree  he  removed  the  locus  will  be  symmetrical 
with  respect  to  the  new  origin  (Theorem  V,  p.  66). 

If  A=  B2  —  4.4(7=0,  equations  (6)  and  (7)  may  still  be  solved  for  h  and  k 

2A       B       D 

if  (Theorem  IV,  p.  81)  —  -  =  —^  =  —  >  when  the  new  origin  (h,  k)  may  be  any 
x>         2  C       lit 

point  on  the  line  2Ax  +  By  +  D  =  0.     In  this  case  every  point  on  that  line  will 
be  a  center  of  symmetry. 

For  example,  consider  «2  +  4a;y  +  4?/2  +  4a;  +  8?/  +  3  =  0.  For  this  equation 
equations  (6)  and  (7)  become 


In  these  equations  the  coefficients  are  all  proportional  and  there  is  an  infinite 
number  of  solutions.  One  solution  is  h  =  —  2,  k  =  0.  For  these  values  the  given 
equation  reduces  to 


The  locus  consists  of  two  parallel  lines  and  evidently  is  symmetrical  with 
respect  to  any  point  on  the  line  midway  between  those  lines. 


148  ANALYTIC   GEOMETRY 

MISCELLANEOUS  PROBLEMS 

1.  Simplify  and  plot. 

(a)  y2  -  5  y  +  6  =  0.  .;(e)  x2  +  ±xy  +  y2  =  8. 

-'  (b)  x2  +  2xy  +  y2  -  6x  -  Qy  +  5  =  0.     (f)  x2  -  9y2  -  2x  -  36y  +  4  =  0. 
(c)  7/2  +  6x-10y  +  2  =  0.  (g)  25t/2-16x2  +  50y-119  =  0. 

_J(d)  x2  +  4y2-  8x-16y  =  0.  (h)  x2  -f  2xy  +  y2  -  8x  =  0. 

2.  Find  the  point  to  which  the  origin  must  be  moved  to  remove  the  terms 
of  the  first  degree  from  an  equation  of  the  second  degree  (Theorem  VII). 

3.  To  what  point  (^,  k)  must  we  translate  the  axes  to  transform 

(1  -  e2)  x2  +  y2  —  2px  +  p2  =  0  into  (1  -  e2)  x2  +  y2  —  2  e^px  -  e*p~  =  0  ? 

4.  Simplify  the  second  equation  in  problem  3. 

5.  Derive  from  a  figure  the  equations  for  rotating  the  axes  through  -f  - 

Tt  ~ 

and  --  5  and  verify  by  substitution  in  (II),  p.  138. 
2 

6.  Prove  thnt  every  equation  of  the  first  degree  may  be  transformed  into 
y'  =  0  by  moving  the  axes.     In  how  many  ways  is  this  possible  ? 

7.  The  equation  for  rotating  the  polar  axis  through   an   angle  <f>  is 
0  =  6'+<}>. 

8.  The  equations  of  transformation  from  rectangular  to  polar  coordi- 
nates, when  the  pole  is  the  point  (A,  k)  and  the  polar  axis  makes  an  angle  of 
<f>  with  the  X-axis,  are 

x  =  h  +  pcos(0  +  0), 
y  —  k  +  p  sin  (0  +  0). 

9.  The  equations  of  transformation  from  rectangular  coordinates  to 
oblique  coordinates  are 

x  =  x'  +  y'  cos  w, 
y  =  y'  sin  o>, 

if  the  JT-axes  coincide  and  the  angle  between  OX.'  and  OY'  is  w. 

10.  The  equations  of  transformation  from  one  set  of  oblique  axes  to  any 
other  set  with  the  same  origin  are 

,  sin  (w  —  0)         .  sin  (w  —  \f/) 
x  =  x  —  J  --  h  y  -    -  > 
sin  a)  BID.  w 


sin  (a          sm  w 

where  w  is  the  angle  between  OX  and  OF,  0  is  the  angle  from  OX  to  OX', 
and  ^  is  the  angle  from  OX  to  OY'. 


CHAPTER  VIII 


CONIC  SECTIONS  AND  EQUATIONS  OF  THE  SECOND  DEGREE 

67.  Equation  in  polar  coordinates.  The  locus  of  a  point  P  is 
called  a  conic  section*  if  the  ratio  of  its  distances  from  a  fixed 
point  'F  and  a  fixed  line  DD  is  constant.  F  is  called  the  focus, 
DD  the  directrix,  and  the  constant  ratio  the  eccentricity.  The 
line  through  the  focus  perpendicular  to  the  directrix  is  called 
the  principal  axis. 

Theorem  I.  If  the  pole  is  the  focus  and  the  polar  axis  the  princi-' 
pal  axis  of  a  conic  section,  then  the  polar  equation  of  the  conic  is 


(I) 


ep 


1  —  ecosO 


where  e  is  the  eccentricity  and  p  is  the  distance  from  the  directrix 
to  the  focus. 

Proof.    Let  P  be  any  point  on  the  conic.     Then,  by  definition, 


FP 


_ 
~ 


EP 

From  the  figure,  FP  =  p 
and  EP  =  HM  =  p  -f  p  cos  0. 


Substituting  these  values  of  FP  and     n 
EP,  we  have 


or,  solving  for  p, 


p 


cos  6 


1  —  6  COS  0 


Q.E.D. 


*  Because  these  curves  may  be  regarded  as  the  intersections  of  a  cone  of  revolution 
with  a  plane. 

149 


150  ANALYTIC  GEOMETRY 

From  (I)  we  see  that 

1.  A  conic  is  symmetrical  with  respect  to  the  principal  axis. 

For  substituting  —  e  f  or  6  changes  only  the  form  of  the  equation,  since 
cos  (—  6)  =  cos  0. 

2.  In  plotting,  no  values  of  0  need  be  excluded. 

The  other  properties  to  be  discussed  (p.  127)  show  that  three 
cases  must  be  considered  according  as  e  =  1. 
The  parabola  e  =  1.    When  e  =  1,  (I)  becomes 

p 

~  1  -  cos  0' 
and  the  locus  is  called  a  parabola. 

1.  For   0  =  0  p  =  oo,    and    for   0  =  ?r    p="|-     The   parabola 
therefore   crosses   the  principal  axis  but  once  at  the  point   O, 

called  the  vertex,  which  is  ^  to  the  left  of  the  focus  F}  or  mid- 
way between  F  and  DD. 

2.  p  becomes  infinite  when  the  denominator,  1  —  cos  0,  vanishes. 
If  1  —  cos  0  =  0,  then  cos  0  =  1;   and  hence  0  =  0  is  the  only 
value  less  than  2  it  for  which  p  is  infinite. 

TT 

3.  When  0  increases  from  0  to  —  ? 
D 

then        cos  0  decreases  from  1  to  0, 

1  —  cos  0  increases  from  0  to  1, 
p  decreases  from  oo  to  p, 
and  the  point  P  (p,  0)  describes  the  parabola 
from  infinity  to  B. 

When  0  increases  from  —  to  TT, 

then  cos  0  decreases  from  0  to  —  1, 

1  —  cos  0  increases  from  1  to  2,         • 

p  decreases  from  p  to  ^? 
& 

and   the  point  P  (p,  0)   describes  the  parabola  from  B  to  the 
vertex  0. 


CONIC  SECTIONS 


151 


On  account  of  the  symmetry  with  respect  to  the  axis,  when 

3  TT 

0  increases  from  TT  to  -^-  >  P  (p,  0)  describes  the  parabola  from  0 

,, 

to  B*\  and  when  6  increases  from  -— -  to  2  TT,  from  13'  to  infinity. 

a 

When  e  <  1  the  conic  is  called  an  ellipse,  and  when  e  >  1, 
an  hyperbola.  The  points  of  similarity  and  difference  in  these 
curves  are  brought  out  by  considering  them  simultaneously. 


The  ellipse,  e<\. 
ep 


1.  For  e  =  0  p  — 


The  hyperbola,  e  >  1. 

ep  e 


1  - 


l-e 


1.  For  0  =  0  p  = 


l-e      l-e 


As  e  <  1,  the  denominator,  and  hence      As  e  >  1,  the  denominator,  and  hence 
p,  is  positive,  so  that  we  obtain  a  point      p,  is  negative,  so  that  we  obtain  a 

point  A  on  the  hyperbola  to  the  left 

ofF. 


A  on  the  ellipse  to  the  right  of  F. 

As  -^—  =  I  when  e<l,  according  as  e 

•>  1      l~e< 

jfe  - ,  then  FA  may  be  greater,  equal  to,  or 

less  than  FH. 


As  —  —  >1  (numerically)  when  e>l, 
l-e 


then 


;  so  A  lies  to  the  left  of  H. 


J 
1-fe 


p.    pis  For0  =  7f/>  = 


ep 


p.    pis 


1+e      1  +  e 

positive,  and  hence  we  obtain  a  point      positive,  and  hence  we  obtain  a  sec- 
ond point  A'  to  the  left  of  F. 


A'  to  the  left  of  F. 

As  -— <1,  then  p<p;    BO  A"  lies 

1  +  e 
between  H  and  F. 

A  and  ^.'  are  called  the  vertices  of 

the  ellipse. 


As  — —  <1.   then  p<p:    so   A'   lies 
1  +  e 

between  H  and  J^. 

-4  and  A'  are  called  the  vertices  of 
the  hyperbola. 


152 


ANALYTIC  GEOMETRY 


The  ellipse,  e  <  1. 

2.  p  becomes  infinite  if 

1  —  e  cos  e  =  0, 

or  cos  e  =  -  • 

e 

As  e<l,  then  ->1;   and  hence 

£ 

there  are  no  values  of  0  for  which 
p  becomes  infinite. 

3.  When 

0  increases  from  0  to  — , 

then  cos  0  decreases  from  1  to  0, 
1  —  e  cos  0  increases  from  1  —  e  to  1 ; 
hence  p  decreases  from  —  —  to  ep, 

L  —  & 

and  P  (p,  0)  describes  the  ellipse  from 
A  to  C. 


"When  0  increases  from  —  to  it, 

2 

then  cos  0  decreases  from  0  to  —  1, 
1  —  e  cos  0  increases  from  1  to  1  +  e ; 

hence       p  decreases  from  ep  to  — — , 

1  -(-  6 

and  P  (p,  0)  describes  the  ellipse  from 
C  to  A'. 

The  rest  of  the  ellipse,  A'C'A, 
may  be  obtained  from  the  symmetry 
with  respect  to  the  principal  axis. 

The  ellipse  is  a  dosed  curve. 


The  hyperbola,  e>l. 

2.  p  becomes  infinite  if 

1  —  e  cos  0  =  0, 

or  cos  0  =  -  - 

e 

As  e>l,  then  -<1;  and  hence 

e 

there  are  two  values  of  0  for  which 
p  becomes  infinite. 

3.  When  l 

0  increases  from  0  to  cos-1  (  -  )» 

then  cos  0  decreases  from  1  to  - » 

e 

l  —  ecos0  increases  from  1  -  e  to  0  ; 


hence 


ep 
p  decreases  from to  —  oo, 


and  P(p,  0)  describes  the  lower  half 
of  the  left-hand  branch  from  A  to 
infinity. 
When 

(1  \         7f 
-   I  tO—  > 
e/      2 


then 


cos  0  decreases  from  -  to  0, 
e 

1  —  e  cos  0  increases  from  0  to  1 ; 
hence          p  decreases  from  oo  to  ep, 
and  P  (p,  0)  describes  the  upper  part  of 
the  right-hand  branch  from  infinity 

to  a 

When  0  increases  from  —  to  TT, 

then  cos  0  decreases  from  0  to  —  1, 
1  —  e  cos  0  increases  from  1  to  1 4-  e ; 

hence       p  decreases  from  ep  to  — — » 

1+e 
and  P  (p,  0)  describes  the  hyperbola 

from  C  to  A'. 

The  rest  of  the  hyperbola,  A'C' 
to  infinity  and  infinity  to  A,  may  be 
obtained  from  the  symmetry  with 
respect  to  the  principal  axis. 

The  hyperbola  has  two  infinite 
branches. 


CONIC  SECTIONS 


153 


PROBLEMS 

1.  Plot  and  discuss  the  following  conies.     Find  e  and  p,  and  draw  the 
focus  and  directrix  of  each. 

W,  =  -^—  <e) 

i  -  cos  e 

2 


3  -  cos  e 


1  -  2  cos  0 


2  -  cos  9 


2  - 


3  -  4  cos  0  • 

2.  Transform  the  equations  in  problem  1  into  rectangular  coordinates, 
simplify  by  the  Rule  on  p.  141,  and  discuss  the  resulting  equations.  Find 
the  coordinates  of  the  focus  and  the  equation  of  the  directrix  in  the  new 
variables.  Plot  the  locus  of  each  equation,  its  focus,  and  directrix  on  the 
new  axes. 

Ana.   (a)  y2  =  4z,  (1,0),  x  =  -  1. 


, 


_ 

-' 


(d)  2/2  =  6x,  (|, 


27 


S.  Transform  (I)  into  rectangular  coordinates,  simplify,  and  find  the  coor- 
dinates of  the  focus  and  the  equation  of  the  directrix  in  the  new  rectangular 
coordinates  if  (a)  e  =  1,  (b) 

(a)  ?/  = 


r2 


(1  - 


?y2  /          C20)  \  V 

-ZL.B1.    (  --^-,   0),   X  =  -- 
22  'V         1  _  e2          )  1  _ 


l-e» 


154  ANALYTIC   GEOMETRY 

4.  Derive  the  equation  of  a  conic  section  when  (a)  the  focus  lies  to  the 
left  of  the  directrix  ;  (b)  the  polar  axis  is  parallel  to  the  directrix. 


Ans.  (a),=      ,  :   (b)  ,  = 


1  -f  e  cos  6  I  -  e  sin  6 

5.  Plot  and  discuss  the  following  conies.     Find  e  and  p,  and  draw  the 
directrix  of  each. 

(a)  P  =  1  +  cos*'  (C)  P  =  3  +  10 cose' 

(b)  p  =  — ?— .  (d)  p  = 


1  -  sin  B  3  -  si 

68.  Transformation  to  rectangular  coordinates. 

Theorem  II.  If  the  origin  is  the  focus  and  the  X-axis  the  princi- 
pal axis  of  a  conic  section,  then  its  equation  is 

where  e  is  the  eccentricity  and  x  =  —  p  is  the  equation  of  the 
directrix. 

Proof.    Clearing  fractions  in  (I),  p.  149,  we  obtain 

p  —  ep  cos  0  =  ep. 
Set  p  =  ±  Vx2  -f-  y2  and  p  cos  0  —  x  (p.  131).     This  gives 

±  ~vx'2  -f-  y2  —  ex  =  ep, 
or  i  v  x2  -J-  y2  =  ex  -J-  ep. 

Squaring  and  collecting  like  powers  of  x  and  y,  we  have  the 
required  equation.  Since  the  directrix  DD  (Fig.,  p.  149)  lies  p 
units  to  the  left  of  F  its  equation  is  x  =—p.  Q.E.D. 

69.  Simplification  and  discussion  of  the  equation  in  rectangu- 
lar coordinates.     The  parabola,  e  =  1. 

When  e  =  1,  (II)  becomes 

Applying  the  Rule  on  p.  141,  we  substitute 

(1)  x  =  x1  +  h,  y  =  y'  4-  k, 
obtaining 

(2)  y'2  -  2px'  +  2ky'  +  k2  -  2ph  -pz  =  Q. 


CONIC  SECTIONS 


155 


Set  the  coefficient  of  y'  and  the  constant  term  equal  to  zero 
solve  for  h  and  k.     This  gives 


Substituting  these  values  in  (2)  and  dropping  primes,  the  equa- 
tion of  the  parabola  becomes  y2  =  2px. 

From  (3)  we  see  that  the  origin  has  been 
removed  from  F  to  0,  the  vertex  of  the    D 
parabola.     It  is  easily  seen  that  the  new 


coordinates  of  the  focus  are  [  £,  0  ) ,  and    x' 

V      I 
the    new   equation    of    the    directrix    is 


-- i- 


D 


Hence 


Theorem  III.    If  the  origin  is  the  vertex  and  the  X-axis  the  axis 
of  a  parabola,  then  its  equation  is 

(III)  y«  : 


The  focus  is  the  point  (  — ,  0  I,  and  the  equation  of  the  directrix 
_p  V     / 

2' 

A  general  discussion  of  (III)  gives  us  the  following  properties  of  the 
parabola  in  addition  to  those  already  obtained 
(p.  150). 

1.  It  passes  through  the  origin  but  does  not  cut 
the  axes  elsewhere. 

2.  Values  of  x  having  the  sign  opposite  to  that 
of  p  are  to  be  excluded  (Rule,  p.  66).     Hence  tl:o 
curve  lies  to  the  right  of  YY'  when  p  is  positive  ar.d 
to  the  left  when  p  is  negative. 

3.  No  values  of  y  are  to  be  excluded ;  hence  this- 
curve  extends  indefinitely  up  and  down. 

Theorem  IV.    If  the  origin  is  the  vertex  and  the  Y-axis  the  axis 
of  a  parabola,  then  its  equation  is 


(IV) 


=  2py. 


\ 


156 


ANALYTIC  GEOMETRY 


/    p\ 

The  focus  is  the  point  (  0,  -  } ,  and  the  equation  of  the  directrix 

A  5  P 


Proof.  Transform  (III)  by  rotating  the 
axes  through  —  — .   Equations  (II),  p.  138, 

give  us  for  0  =  — 


x  =  y , 

y  =  —  x>- 

Substituting  in  (III)  and  dropping  primes,  we  obtain  x*  =  2py. 

Q.E.D. 

Y' 

turned  through  ^  in  the  positive  direction. 

The  parabola  lies  above  or  below  the  X-axis 
according  as  p  is  positive  or  negative. 

Equations  (III)  and  (IV)  are  called 
the  typical  forms  of  the  equation  of  the 
parabola. 

Equations  of  the  forms 

Ax2  -f  Ey  =  0  and  Cy2  -f-  Dx  =  0, 


After  rotating  the  axes  the  whole  figure  is 


where  A,  E,  C,  and  D  are  different  from  zero,  may,  by  transpo- 
sition and  division,  be  written  in  one 
of  the  typical  forms  (III)  or  (IV), 
so  that  in  each  case  the  locus  is  a 
-    parabola. 

Ex.  1.  Plot  the  locus  of  x2  +  4  y  =  0  and 
find  the  focus  and  directrix. 

Solution.  The  given  equation  may  be 
written 

Comparing  with  (IV),  the  locus  is  seen  to  be  a  parabola  for  which  p  =  -  2. 
Its  focus  is  therefore  the  point  (0,  —  1)  and  its  directrix  the  line  y  —  I. 

Ex.  2.    Find  the  equation  of  the  parabola  whose  vertex  is  the  point  (X 
(3,  -  2)  and  whose  directrix  is  parallel  to  the  F-axis,  if  p  =  3. 


Y 

1 

D 

a 

0 

A" 

X 

•^""~ 

N 

/ 

% 

i-2J 

\ 

/ 

\ 

/ 

5 

i 

/ 

\ 

CONIC  SECTIONS 


157 


Solution.    Referred  to  O'X'  and  O'F'  as  axes,  the  equation  of  the  parabola 
(Theorem  III) 

k  -,,'2  _  A  />.'  '  -O 


O' 


x 


» 
The  equation  for  translating  the  axes  from 
0  to  (y  are  (Theorem  I,  p.  136) 

x  =  x'  +  3,  y  =  y'  -2, 
whence 

(5)  x'  =  x  -  3,  y'  =  y,+  2. 

Substituting  in  (4),  we  iHkin  as  the  re- 
quired equation 

(y +  2)*  =  0(a&-8 

or  y2  -  6  x  +  4  y  -f  22  = 

Referred  to  O'JT'  and  O'F',  the  coordinates 
of  F  are  (Theorem  III)  (|,  0)  and  the  equa- 
tion of  DD  is  x'  =  —  .f .     By  (5)  we  see  that, 
referred  to  OX  and  OF,  the  coordinates  of  .Fare  (f,  —  2)  and  the  equation 
of  DD  is  x  =  |. 

PROBLEMS 

1.  Plot  the  locus  of  the  following  equations.     Draw  the  focus  and  direc- 
trix in  each  case. 

(a)  y2  =  4x.  (d)  3,2  _  63  =  0. 

(b)  2/2  +  4x  =  o.  (e)  z2  +  10  y  =  0. 

(c)  x2-82/  =  0.  (f)  2/2  +  x  =  0. 

2.  If  the  directrix  is  parallel  to  the  F-axis,  find  the  equation  of  the 
parabola  for  which 

(a)  p  =  6,  if  the  vertex  is  (3,  4). 

(b)  p  =  -  4,  if  the  vertex  is  (2,  -  3). 

(c)  p  =  8,  if  the  vertex  is  (-  5,  7). 

(d)  p  =  4,  if  the  vertex  is  (h,  k). 

3.  The  chord  through  the  focus  perpendicular  to  the  axis  is  called  the  latus 
rectum.      Find  the  length  of  the  latus  rectum  of  y2  =  2px.  Ans.    2 p. 

4.  What  is  the  equation  of  the  parabola  whose  axis  is  parallel  to  the  axis 
of  y  and  whose  vertex  is  the  point  (a,  /S)  ?        -4ns.    (x  —  a)2  =  2p  (y  —  /3). 

5.  Transform  to  polar  coordinates  and  discuss  the  resulting  equations 
(a)  y2  =  2px,  (b)  x2  =  2py. 

6.  Prove  that  the  abscissas  of  two  points  on  the  parabola  (III)  are  propor- 
tjonal  to  the  squares  of  the  ordinates  of  those  points, 


Ans.  (y  -  4)2  =  12  (x  -  3). 

Ans.  (2/  +  3)2  =  -8(x-2) 

Ans.  (y-7)2  =  16(x  +  5). 

Ans.  y-k*  =  Sx-h. 


158 


ANALYTIC   GEOMETRY 


70.  Simplification  and  discussion  of  the  equation  in  rectan- 
gular coordinates.  Central  cpnics,  e  J  1.  When  e  ^  1,  equation 
(II),  p.  154,  is 

(1  -  e2)x2  +  y2-2  e2px  -  e2p2  =  0. 

To  simplify  (Kule,  p.  141),  set 
which  gives 


-  2  e'2p 

-  2  e2pk 
-e2p2 

Setting  the  coefficients  of  x'  and  y'  equal  to  zero  gives 
2  h  (1  -  e2)  -  2  e2p  =  0,  2  k  =  0, 


whence 
(3) 


h  = 


k  =  0. 


Substituting  in  (2)  and  dropping  primes,  we  obtain 


or 


^2         g2^2 


=  1. 


This  is  obtained  by  transposing  the  constant  term,  dividing  by  it,  and  then 
dividing  numerator  and  denominator  of  the  first  fraction  by  1  —  e2. 


The  ellipse,  e  <  1. 

From  (3)  it  is  seen  that  h  is  posi- 
tive when  e  <l.  Hence  the  new  ori- 
gin O  lies  to  the  right  of  the  focus  F. 


The  hyperbola,  e>l. 

From  (3)  it  is  seen  that  h  is  nega- 
tive when  e  >  1.  Hence  the  new  ori- 
gin O  lies  to  the  left  of  the  focus  F. 

e2 

Further,  >  1  numerically,  so 

1  -  e2 

h  >  p  numerically  ;  and  hence  the 
new  origin  lies  to  the  left  of  the 
directrix  DD. 


CONIC  SECTIONS 


159 


The  locus  of  (4)  is  symmetrical  with  respect  to  YY'  (Theorem 
V,  p.  66).     Hence  0  is  the  middle  point  of  A  A'.     Construct  in 


either  figure  F'  and  D'D'  symmetrical  respectively  to  F  and  DD 
with  respect  to  YY'.  Then  F'  and  D'D'  are  a  new  focus  and 
directrix. 

For  let  P  and  Pf  be  two  points  on  the  curve,  symmetrical  with  respect  to  FF7. 
Then  from  the  symmetry  PF  =  P'F'  and  PE  =  P'E'.  But  since,  by  definition, 

PF  P'F' 

-=—  -  =  e,  then  =  e.    Hence  the  same  conic  is  traced  by  P',  using  F'  as  focus 

IT  Hi  P  Jit 

and  D'D'  as  directrix,  as  is  traced  by  P,  using  F  as  focus  and  DD  as  directrix. 

Since  the  locus  of  (4)  is  symmetrical  with  respect  to  the  origin 
(Theorem  V,  p.  66),  it  is  called  a  central  conic,  and  the  center  of 
symmetry  is  called  the  center.  Hence  a  central  conic  has  two  foci 
and  two  directrices. 

The  co6rdiriates  of  the  focus  F  in  either  figure  are 


For  the  old  coordinates  of  F  were  (0,  0).     Substituting  in  (1),  the  new  coordi- 
nates are  x  =  -  h,  y'  =  -  k,  or,  from  (3),  (  -    -^—^  0  V 


The  coordinates  of  F'  are  therefore 


rM- 

1  —  e2      I 


t) 

The  new  equation  of  the  directrix  DD  is  x  =  —  ^ — ^ 

JL  —  e 


160 


ANALYTIC  GEOMETRY 


For   from    (1)    and    (3),   x  = 


1-e2 
(Theorem  II)  and  dropping  primes,  we  obtain  x  =  — 


=  y'.     Substituting  in  x  =  —  p 
P 


P 
Hence  the  equation  of  D'D'  is  x  =     _    2- 

We  thus  have  the 

Lemma.    The  equation  of  a  central  conic  whose  center  is  the  origin 
and  whose  principal  axis  is  the  X-axis  is 

T2 

W  — r^  +  - 

^      '  ^**/v-k-«  n 


=  1. 


(1  _ 


1-e 


Its  foci  are  the  points 


(  ±  -  —  —  2>  0  ) 
\     1  —  e       I 


and  its  directrices  are  the  lines  x  =  ± 


The  ellipse,  e<\. 
For  convenience  set 
ep  _e*~ 


(5)  a  = 


1-e2 


I-. 


c  = 


«2  and  62  are  the  denominators  in  (4) 
and  c  is  the  abscissa  of  one  focus.  Since 
e<l,  1-e2  is  positive;  and  hence  a,  ft2, 
and  c  are  positive. 

We  have  at  once 

*,2-n2  *>2-n2 

a2  -  62  = 


(1  -  e2)2      1.- 


1-e2 

TAe  hyperbola,  e>l. 
For  convenience  set 

(6) 
o=  — 


ep 


1-e2 


1-e2 


a2  and  -  62  are  the  denominators  in  (4) 
and  c  is  the  abscissa  of  one  focus.  Siuce 
e  >  1, 1-  e*  is  negative;  and  hence  a,  62,  rnd 
c  arepositfiue. 

We  have  at  once 

,,9^9  ^5^2 

a2  +  ft2  = 


_  e2) 


and 
a2 


(1  -  e2) 


and 
a2  _ 
7  ~ 


(1  - 


1-e2 


c       (1-e2)2      1-e2      1-e2 

Hence  the  directrices  (Lemma)  are  Hence  the  directrices  (Lemma)  are 


a2 

the  lines  x  =  ±  —  • 
c 


the  lines  x  =  ±  —  • 


By  substitution  from  (5)  in  (4)  we  By  substitution  from  (6)  in  (4)  we 


obtain 


obtain 


A* 

CONIC  SECTIONS 


161 


I  The  ellipse,  e  <  1. 

The  intercepts  are  x  =  ±  a  and 
2/  =  ±  6.  -4.4'  =  2  a  is  called  the 
major  axis  and  BB'  =  2b  the  minor 
axis.  Since  a2  —  b2  =  c2  is  positive, 
then  a  >  6,  and  the  major  axis  is 
greater  than  the  minor  axis. 


The  hyperbola,  e>l. 
The  intercepts  are  x  =  ±  a,  but  the 
hyperbola  does  not  cut  the  Y-axis. 
AA'  =  2  a  is  called  the  transverse 
axis  and  BB'  =  26  the  conjugate 
axis. 


Hence  we  may  restate  the  Lemma 
as  follows. 

Theorem  V.  The  equation  of  an 
ellipse  whose  center  is  the  origin  and 
whose  foci  are  on  the  X-axis  is 

:/'2        v/2 

where  2  a  is  the  major  axis  and  2  b  the 
minor  axis.  If  c2  =  az  —  ft2,  then  the 
foci  are  (±  c,  0)  and  the  directrices 

a2 

are  x  =  ±  — 
c 

Equations  (5)  also  enabje  us  to 
express  e  and  p,  the  constants  of  (I), 
p.  149,  in  terms  of  a,  6,  and  c,  the 
constants  of  (V).  For 


Hence  we  may  restate  the  Lemma 
as  follows. 

Theorem  VI.  The  equation  of  an 
hyperbola  whose  center  is  the  origin 
and  whose  foci  are  on  the  X-axis  is 

<vi>     S-S=1- 

where  2  a  is  the  transverse  axis  and  2b 
the  conjugate  axis.  If  c2  =  a2  +  ft2, 
then  the  foci  are  (±  c,  0)  and  the 

a2 

directrices  are  x  =  ±  —  • 
c 

Equations  (6)  also  enable  us  to 
express  e  and  p,  the  constants  of  (I), 
p.  149,  in  terms  of  a,  6,  and  c,  the 
constants  of  (VI).  For 


= 

a 

and 

M  = 
v  '        c 


= 


and 


162 


ANALYTIC  GEOMETRY 


The  ellipse,  e<l. 

In  the  figure  OB  =  b,  OF'  =  c ; 
and  since  c2  =  a2  -  62,  then  BF'  =  a. 
Hence  to  -draw  the  foci,  with  B  as  a 
center  and  radius  OA,  describe  arcs 
cutting  XX'  at  F  and  F'.  Then  F 
and  F'  are  the  foci. 

If  a  =  6,  then  (V)  becomes 

x2  +  y2  =  aat 
whose  locus  is  a  circle. 

Transform   (V)   by   rotating   the 

Tt 

axes  through  an  angle  of (Theo- 
rem II,  p.  138).     We  obtain 

Theorem  VII.  The  equation  of  an 
ellipse  whose  center  is  the  origin  and 
whose  foci  are  on  the  Y-axis  is 


(VII) 


where  2  a  is  the  major  axis  and  2  b  is 
the.  minor  axis.  If  c2  =  a2  -  62,  the 
foci  are  (0,  ±  c)  and  the  directrices 

a2 
are  the  lines  y  =  ±  —  - 


The  hyperbola,  e>\. 

In  the  figure  OB  =  b,  OA'  =  a ; 
and  since  c2  =  a2  +  62,  then  BA'  =  c. 
Hence  to  draw  the  foci,  with  O  as  a 
center  and  radius  BA',  describe  arcs 
cutting  XX'  at  F  and  F'.  Then  F 
and  F'  are  the  foci. 

If  a  =  b,  then  (VI)  becomes 

x2  -  y2  =  a2, 

whose  locus  is  called  an  equilateral 
hyperbola. 

Transform  (VI)  by  rotating  the 

axes  through  an  angle  of (Theo- 
rem II,  p.  138).     We  obtain 

Theorem  VET  The  equation  of  an 
hyperbola  whose  center  is  the  origin 
and  whose  foci  are  on  the  Y-axis  is 

<vm>  -+=l' 


X' 


D' 


D 


where  2  a  is  the  transverse  axis  and  2b 
is  the  conjugate  axis.  Ifc2  =  a2  +  b2, 
the  foci  are  (0,  ±  c)  and  the  directrices 

a2 

are  the  lines  y  =  ±  — . 
c 


CONIC  SECTIONS 


163 


The  ellipse,  e<\. 

The  essential  difference  between 
(V)  and  (VII)  is  that  in  (V)  the  de- 
nominator of  x2  is  larger  than  that 
of  y2,  while  in  (VII)  the  denominator 
of  y*  is  the  larger.  (V)  and  (VII) 
are  called  the  typical  forms  of  the 
equation  of  an  ellipse. 


The  hyperbola,  e>l. 

The  essential  difference  between 
(VI)  and  (VIII)  is  that  the  coeffi- 
cient of  y2  is  negative  in  (VI),  while 
in  (VIII)  the  coefficient  of  x2  is  nega- 
tive. (VI)  and  (VIII)  are  called  the 
typical  forms  of  the  equation  of  an 
hyperbola. 


An  equation  of  the  form 

Ax2  +  Cif  +  F  =  0, 

where  A,  C,  and  F  are  all  different  from  zero,  may  always  be 
written  in  the  form 


By  transposing  the  constant  term  and  then  dividing  by  it,   and  dividing 
numerator  and  denominator  of  the  resulting  fractions  by  A  and  C  respectively. 

The  locus  of  this  equation  will  be 

1.  An  ellipse  if  a  and  ft  are  both  positive,    a2  will  be  equal  to 
the  larger  denominator  and  b'2  to  the  smaller. 

2.  An  hyperbola  if  a.  and  ft  have  opposite  signs,     a2  will  be 
equal  to  the  positive  denominator  and  b2  to  the  negative  denomi- 
nator. 

3.  If  a  and  ft  are  both  negative,  (11)  will  have  no  locus. 

Ex.  1.    Find  the  axes,   foci,  directrices,  and 
eccentricity  of  the  ellipse  4  x2  +  y2  =  16. 

Solution.    Dividing  by  16,  we  obtain 

*+*  =  !. 
4       16 

The  second  denominator  is  the  larger.     By 
comparison  with  (VII), 

W-  =  4,  a2  =  16,  c2  =  16  -  4  =  12. 
Hence  6  =  2,        a  =  4,       c  =  Vl2. 

The  positive  sign  only  is  used  when  we  extract  the 
square  root,  because  a,  b,  and  c  are  essentially  positive. 


164  ANALYTIC  GEOMETRY 

Hence  the  major  ?aaa_AA'  =  8,  the  minor  axis  BB'  =  4,  the  foci  F  and  F' 
are  the  points  (0,  ±  "^12),  and  the  equations  of  the  directrices  DD  and  D'l/ 

a2  16  4    _ 

are  y  =  ±  —  =  ±  — —  -  ±  -' Vl2. 
c  Vl2 

Vl2  4         1 

From  (7)  and  (9),  e  =  -  -  and  p  =  — —  =  -  Vl2. 
4  V12      3 


PROBLEMS 

1 .  Plot  the  loci,  directrices,  and  foci  of  the  following  equations  and  find 

=  81.  (e)  9y2  -  4x2  =  36. 

(b)  9  x2  -  16  y2  =  144.  (f )  x2  -  y2  =  25. 

(c)  16  x2  +  y2  =  25.  (g)  4  x2  +  7  y2  =  13. 

(d)  4x2  +  9y2  =  36.  (h)  5x2  -  3y2  =  14. 

2.  Find  the  equation  of  the  ellipse  whose  center  is  the  origin  and  whose 
foci  are  on  the  .X-axis  if 

(a)  a  =  5,  b  =  3.  Ans.  9x2  +  25  y2  =  225. 

(b)  a  =  6,  e  =  $.  Ans.  32  x2  +  36  y2  =  1152. 

(c)  6  =  4,  c  =  3.  Ans.  16  x2  +  25  y2  =  400. 

(d)  c  =  8,  e  =  f.  4ns.  5x2  +  9y2  =  720. 

3.  Find  the  equation  of  the  hyperbola  whose  center  is  the  origin  and 
whose  foci  are  on  the  JT-axis  if 

(a)  a  =  3,  6  =  5.  Ans.  25 x2  -  9y2  =  225. 

(b)a  =  4,  c  =  5.  Ans.  9x2  —  16  y2  =  144. 

(c)  e  =  f ,  a  =  5.  Ans.  5  x2  -  4  y2  =  125. 

(d)  c  =  8,  e  =  4.  4ns.  15x2  -  y2  =  60. 

4.  Show  that  the  latus  rectum  (chord  through  the  focus  perpendicular  to 

252 

the  principal  axis)  of  the  ellipse  and  hyperbola  is . 

a 

5.  What  is  the  eccentricity  of  an  equilateral  hyperbola  ?        Ans.    V2. 

6.  Transform  (V)  and  (VI)  to  polar  coordinates  and  discuss  the  resulting 
equations. 

7.  Where  are  the  foci  and  directrices  of  the  circle  ? 

8.  What  are  the  equations  of  the  ellipse  and  hyperbola  whose  centers 
are  the  point  (a,  /3)  and  whose  principal  axes  are  parallel  to  the  X-axis? 

* -aO2  ,   (if  ~  W  _  -.  .  (*  ~  <*)*      (V  -  W      , 
~~  +  ~~-l-~ ~ 


CONIC  SECTIONS  165 

71.  Conjugate  hyperbolas  and  asymptotes.  Two  hyperbolas 
are  called  conjugate  hyperbolas  if  the  transverse  and  conjugate 
axes  of  one  are  respectively  the  conjugate  and  transverse  axes  of 
the  other.  They  will  have  the  same  center  and  their  principal 
axes  (p.  149)  will  be  perpendicular. 

If  the  equation  of  an  hyperbola  is  given  in  typical  form,  then 
the  equation  of  the  conjugate  hyperbola  is  found  by  changing  the 
signs  of  the  coefficients  of  x2  and  y*  in  the  given  equation. 

For  if  one  equation  be  written  in  the  form  (VI)  and  the  other  in  the  form  (VIII), 
then  the  positive  denominator  of  either  is  numerically  the  same  as  the  negative 
denominator  of  the  other.  Hence  the  transverse  axis  of  either  is  the  conjugate 
axis  of  the  other. 

Thus  the  loci  of  the  equations 

(1)  16z2  -  ?/2  =  16  and  -I6x*  +  y*=  16 

are  conjugate  hyperbolas.    They  may  be  written 

a;2      y2  X2      y2 

___  =  land__  +  ^  =  1. 

The  foci  of  the  first  are  on  the  Jf-axis,  those  of  the  second  on  the  F-axis.  The 
transverse  axis  of  the  first  and  the  conjugate  axis  of  the  second  are  equal  to  2, 
while  the  conjugate  axis  of  the  first  and  the  transverse  axis  of  the  second  are 
equal  to  8. 

The  foci  of  two  conjugate  hyperbolas  are  equally  distant  from 
the  origin. 

For  c2  (Theorems  VI  and  VIII)  equals  the  sum  of  the  squares  of  the  semi- 
transverse  and  semi-conjugate  axes,  and  that  sum  is  the  same  for  two  conjugate 
hyperbolas. 

Thus  in  the  first  of  the  hyperbolas  above  c2  =  1  +  16,  while  in  the  second 


If  in  one  of  the  typical  forms  of  the  equation  of  an  hyperbola 
we  replace  the  constant  term  by  zero,  then  the  locus  of  the  new 
equation  is  a  pair  of  lines  (Theorem,  p.  59)  which  are  called  the 
asymptotes  of  the  hyperbola. 

Thus  the  asymptotes  of  the  hyperbola 

(2)  bV  -  aY  =  <*?& 
'are  the  lines 

(3)  bW  -  aY  =  0, 
or 

(4)  bx  -f  ay  =  0  and  bx  —  ay  =  0. 

0 


166  ANALYTIC  GEOMETRY 

Both  of  these  lines  pass  through  the  origin,  and  their  slopes  are  respectively 
(5)  --and-- 

An  important  property  of  the  asymptotes  is  given  by 

Theorem  IX.  The  branches  of  the  hyperbola  approach  its  asymp- 
totes as  they  recede  to  infinity. 

Proof.  Let  P±  (xly  y-^)  be  a  point  on  either  branch  of  (2)  near 
the  first  of  the  asymptotes  (4).  The  distance  from  this  line  to 
P^Fig.,  p.  167)  is  (Rule,  p.  97) 

x/>>.  T bxl  -\-  ay i 

+  V&2  -f  a2 
Since  Px  lies  on  (2),    b^x^  —  ahj^  = 

Factoring,  bxl  +  ay±  =  — 


Substituting  in  (6),     d  = 


+  a2  bx±  — 


As  PI  recedes  to  infinity,  x±  and  y±  become  infinite  and  d 
approaches  zero. 

For  bxi  and  ay^  cannot  cancel,  since  Xi  and  y±  have  opposite  signs  in  the  second 
and  fourth  quadrants. 

Hence  the  curve  approaches  closer  and  closer  to  its  asymptotes. 

Q.E.D. 

Two  conjugate  hyperbolas  have  the  same  asymptotes. 

For  if  we  replace  the  constant  term  in  both  equations  by  zero,  the  resulting 
equations  differ  only  in  form  and  hence  have  the  same  loci. 

Thus  the  asymptotes  of  the  conjugate  hyperbolas  (1)  are  respectively  the  loci  of 

16a;2  _  yz  -  o  and  -!Qx*  +  y2  =  0, 
which  are  the  same. 

An  hyperbola  may  be  drawn  with  fair  accuracy  by  the  fol- 
lowing 

Construction.  Lay  off  OA  =  OA  '  =  a  on  the  axis  on  which  the 
foci  lie,  and  OB  =  OB'  =  b  on  the  other  axis.  Draw  lines  through 
A,  A1,  B,  B'  parallel  to  the  axes,  forming  a  rectangle.*  Draw  the 

*  An  ellipse  may  be  drawn  with  fair  accuracy  by  inscribing  it  in  such  a  rectangle. 


CONIC  SECTIONS 


167 


diagonals  of  the  rectangle  and  the  circumscribed  circle.  Draw 
the  branches  of  the 
hyperbola  tangent  to 
the  sides  of  the  rec- 
tangle at  A  and  A' 
and  approaching  nearer 
and  nearer  to  the  di- 
agonals. The  conju- 
gate hyperbola  may 
be  drawn  tangent  to 
the  sides  of  the  rec- 
tangle at  B  and  B' 
and  approaching  the  diagonals.  The  foci  of  both  are  the  points 
in  which  the  circle  cuts  the  axes. 

The  diagonals  will  be  the  asymptotes,  because  two  of  the  vertices  of  the  rec- 
tangle ( i  a,  ±  6)  will  lie  on  each  asymptote  (4) .  Half  the  diagonal  will  equal  c, 
the  distance  from  the  origin  to  the' foci,  because  c2=  a2  +  62. 

72 .  The  equilateral  hyperbola  referred  to  its  asymptotes.    The  equation 
of  the  equilateral  hyperbola  (p.  162)  is 
(1)  x2  -  y2  =  a2. 

Its  asymptotes  are  the  lines 

x  —  y  =  0  and  x  -f  y  =  0. 

These  lines  are  perpendicular  (Corollary  III,  p.  78),  and  hence  they  may 
be  used  as  coordinate  axes. 

Theorem  X.    The  equation  of  an  equilateral  hyperbola  referred  to  its  asymp- 
totes is 
(X)  2jcy  =  az. 

Proof.    The  axes  must  be  rotated  through 

to  coincide  with  the  asymptotes. 

Hence  we  substitute  (Theorem  II,  p.  138) 

—  X'  -}-  y/ 


X  = 


V2 
in  (1).     This  gives 

(z'  +  y')a      (- 
2 


y  = 


Or,  reducing  and  dropping  primes, 

2xy  = 


Q.E.D. 


168 


ANALYTIC  GEOMETRY 


73.  Focal  property  of  central  conies.  A  line  joining  a  point  on 
a  conic  to  a  focus  is  called  a  focal  radius.  Two  focal  radii,  one  to 
each  focus,  may  evidently  be  drawn  from  any  point  on  a  central 
conic. 


Theorem  XI.  The  sum  of  the  focal 
radii  from  any  point  on  an  ellipse  is 
equal  to  the  major  axis  2  a. 


Proof.    Let  P  be  any  point  on  the 
ellipse.     By  definition  (p.  149), 

r  =  e-PE,  r'  =  e-  PE'. 
Hence  r  +  r7  =  e  (PE  +  PE') 
=  e-HH'. 

From  (7),  p.  161,  e  =  - , 
a 

and  from  the  equations  of  the  direc- 
trices (Theorem  V), 

a2 

c 

Hence  r  +  r/  =  --2  —  =  2o. 

Q.E.D. 


Theorem  Xn.  The  difference  of  the 
focal  radii  from  any  point  on  an 
hyperbola  is  equal  to  the  transverse 
axis  2  a. 


Proof.    Let  P  be  any  point  on  the 
hyperbola.     By  definition  (p.  149), 

r  =  e-PE,  r'  =  e  •  PE'. 
Hence  r'  -  r  —  e  (PW  —  PE) 
=  e-  HH'. 

From  (8),  p.  161,  e=  -, 

and  from  the  equations  of  the  direc- 
trices (Theorem  VI), 


Q.E.D. 


Hence  r'-r  =  --2-  = 
a       c 


74.  Mechanical  construction  of  conies.  Theorems  XI  and  XII  afford  simple 
methods  of  drawing  ellipses  and  hyperbolas.  Place  two  tacks  in  the  drawing 
board  at  the  foci  F  and  F'  and  wind  a  string  about  them  as  indicated. 

If  the  string  be  held  fast  at  A,  and  a  pencil  be  placed  in  the  loop  FPF' 
and  be  moved  so  as  to  keep  the  string  taut,  then  PF  +  PFf  is  constant  and 
P  describes  an  ellipse.  If  the  major  axis  is  to  be  2  a,  then  the  length  of  the 
loop  FPF'  must  be  2  a. 


CONIC  SECTIONS 


169 


If  the  pencil  be  tied  to  the  string  at  P,  and  both  strings  be  pulled  hi  or 
let  out  at  A  at  the  same  time,  then  PF  —  PF  will  be  constant  and  P  will 
describe  an  hyperbola.  If  the  transverse  axis  is  to  be  2  a,  the  strings  must 
be  adjusted  at  the  start  so  that  the  difference  between  PF'  and  PF  equals  2  a. 


To  describe  a  parabola,  place  a  right  triangle  with  one  leg  EB  on  the 
directrix  DZ>.  Fasten  one  end  of  a  string  whose  length  is  AE  at  the  focus 
F,  and  the  other  end  to  the  triangle  at  A.  With  a  pencil  at  P  keep  the 
string  taut.  Then  PF  =  PE  ;  and  as  the  triangle  is  moved  along  DD  the 
point  P  will  describe  a  parabola. 

PROBLEMS 

1.  Find  the  equations  of  the  asymptotes  and  hyperbolas  conjugate  to  the 
following  hyperbolas,  and  plot. 

(a)  4z2  -  ?/2  =  36.  (c)  16z2  -  y2  +  64  =  0. 

^Jb)  9  a2  -  25  y2  =  100.  (d)  8  x2  -  16  y*  +  25  =  0. 

2.  Prove  Theorem  IX  for  the  asymptote  which  passes  through  the  first 
and  third  quadrants. 

3.  If  e  and  ef  are  the  eccentricities  of  two  conjugate  hyperbolas,  then 


4.  The  distance  from  an  asymptote  of  an  hyperbola  to  its  foci  is  numer- 
ically equal  to  6. 

5.  The  distance  from  a  line  through  a  focus  of  an  hyperbola,  perpen- 
dicular to  an  asymptote,  to  the  center  is  numerically  equal  to  a. 

6.  The  product  of  the  distances  from  the  asymptotes  to  any  point  on  the 
hyperbola  is  constant. 

7.  The  focal  radius  of  a  point  PI(XI,  yi)  on  the  parabola  y*  =  2px  is 

f-, 


170  T  ANALYTIC  GEOMETRY 

8.  The  focal  radii  of  a  point  PI(ZI,  y{)  on  the  ellipse  62x2  +  a2?/2  =  a262 
are  r  =  a  —  ex\  and  r'  =  a  +  exi. 

9.  The  focal  radii  of  a  point  on  the  hyperbola  62x2  —  a2?/2  =  a262  are 
r  =  exi  —  a   and   r'  —  e&i  +  a  when  PI  is  on  the   right-hand  branch,  or 
r  =  -  exi  —  a  and  r'  =  -  exi  +  a  when  Pj.  is  on  the  left-hand  branch. 

10.  The  distance  from  a  point  on  an  equilateral  hyperbola  to  the  center 
is  a  mean  proportional  between  the  focal  radii  of  the  point. 

11.  The  eccentricity  of  an  hyperbola  equals  the  secant  of  the  inclination 
of  one  asymptote. 

75.  Types  of  loci  of  equations  of  the  second  degree.    All  of 

the  equations  of  the  conic  sections  that  we  have  considered  are 
of  the  second  degree.  If  the  axes  be  moved  in  any  manner,  the 
equation  will  still  be  of  the  second  degree  (Theorem  IV,  p.  140), 
although  its  form  may  be  altered  considerably.  We  have  now  to 
-consider  the  different  possible  forms  of  loci  of  equations  of  the 
second  degree. 

By  Theorem  VI,  p.  145,  the  term  in  xy  may  be  removed  by 
rotating  the  axes.  Hence  we  only  need  to  consider  an  equation 
of  the  form 

(1)  Ax*  +  Cy*  +  Dx  +  Ey  +  F=0. 

It  is  necessary  to  distinguish  two  cases. 

CASE     I.    Neither  A  nor  C  is  zero. 

CASE  II.    Either  A  or  C  is  zero. 

A  and  C  cannot  both  be  zero,  as  then  (1)  would  not  be  of  the  second  degree. 

Case  I 

When  neither  A  nor  C  is  zero,  then  A  =  B2  —  4  A  C  is  not  zero, 
and  hence  (Theorem  VII,  p.  146)  we  can  remove  the  terms  in 
x  and  y  by  translating  the  axes.  Then  (1)  becomes  (Corollary  I, 
p.  147) 

(2)  Ax12  +  cy2  +  F1  =  0. 

We  distinguish  two  types  of  loci  according  as  A  and  C  have  the 
same  or  different  signs. 


CONIC  SECTIONS 


171 


Elliptic  type,  A  and  C  have  the 
same  sign. 

1.   F/  ^  0.*    Then    (2)    may    be 

written          1 =  1, 


where 


Hence,  if  the  sign  of  F'  is  different 
from  that  of  A  and  C,  the  locus  is  an 
ellipse;  but  if  the  sign  of  F'  is  the 
same  as  that  of  A  and  C,  there  is  no 
locus. 

2.  F'  =  0.  The  locus  is  a  point. 
It  may  be  regarded  as  an  ellipse 
whose  axes  are  zero  and  it  is  called 
a  degenerate  ellipse. 


Hyperbolic  type,  A  and  C  have  dif- 
ferent signs. 

1.   F'  *  0.*    Then    (2)    may    be 

x2      v2 

written  —  +  —  =  1, 

a       0 

'  ' 

where 


Hence  the  locus  is  an  hyperbola  whose 
foci  are  on  the  Y-axis  if  the  signs  of 
F*  and  A  are  the  same,  or  on  the 
X-axis  if  the  signs  of  F'  and  C  are 
the  same. 

2.  F'  =  0.  The  locus  is  a  pair  of 
intersecting  lines.  It  may  be  regarded 
as  an  hyperbola  whose  axes  are  zero 
and  it  is  called  a  degenerate  hyperbola. 


Case  II 

When  either  A  or  C  is  zero  the  locus  is  said  to  belong  to  the 
parabolic  type.  We  can  always  suppose  A  =  0  and  C  =£  0,  so  that 
(1)  becomes 

(3)  Cif  +  Dx  +  Ey  +  F  =  0. 

For  if  A  T±  0  and  (7=0,  (1)  becomes ^z2  +  Dx  +  Ey  +  F=0.  Rotate  the  axes 
(Theorem II,  p.  138)  through  —  by  setting  x=—y',y=  x'.  This  equation  becomes 
Ay  2  +  Ex'  -Dy'  +  F=0,  which  is  of  the  form  (3). 

By  translating  the  axes  (3)  may  be  reduced  to  one  of  the  forms 

(4)  Cif  +  Dx  =  0  or 

(5)  cy  +  Fr  =o. 

For  substitute  in  (3),          x  =  x"  -f  A,  y  =  y'  +  &. 
This  gives 
(6)  C?//2  +  Da/  +  2  C'fc  I  ?/'  +  CW    =0. 


If  we  determine  h  and  k  from 


+  Dh 

+  Ek 
+  F 


then  (6)  reduces  to  (4).    But  if  D  =  0,  we  cannot  solve  the  last  equation  for  h,  so 
that  we  cannot  always  remove  the  constant  term.    In  this  case  (6)  reduces  to  (5). 

*  Read  "  F'  not  equal  to  zero  "  or  "F'  different  from  zero." 


172  ANALYTIC  GEOMETRY 

Comparing  (4)  with  (III),  p.  155,  the  locus  is  seen  to  be  &parab- 

I      ~F' 
ola.     The  locus  of  (5)  is  the  pair  of  parallel  lines  y  =  ±  ^  —  — 

when  F1  and  C  have  different  signs,  or  the  single  line  y  —  0 
when  F'  =  0.     If  F1  and  C  have  the  same  sign,  there  is  no  locus. 
When  the  locus  of  an  equation  of  the  second  degree  is  a  pair  of 
parallel  lines  or  a  single  line  it  is  called  a  degenerate  parabola. 
We  have  thus  proved 

Theorem  XIII.  The  locus  of  an  equation  of  the  second  degree  is 
a  conic,  a  point,  or  a  pair  of  straight  lines,  which  may  be  coincident. 
By  moving  the  axes  its  equation  may  be  reduced  to  one  of  the  three 

forms 

Ax2  +  Cy*  +  F*  =  0,   Cy2  +  Dx  =  0,   Cif  +  F*  =  0, 

where  A  ,  C,  and  D  are  different  from  zero. 

Corollary.    The  locus  of  an  equation  in  which  the  term  in  xy  is 

lacking,  Ax*  +  Cf  +  Dx  +  E?/  +  F  =  0, 

will  belong  to 

the  parabolic  type  if  A  =  0  or  C  •=  0, 

the  elliptic  type  if  A  and  C  have  the  same  sign, 

the  hyperbolic  type  if  A  and  C  have  different  signs. 

PROBLEMS 

1  .  To  what  point  is  the  origin  moved  to  transform  (1)  into  (2)  ? 

/       D  E  \ 

Ans-  -'- 


2.  To  what  point  is  the  origin  moved  to  transform  (3)  into  (4)  ?   into  (5)  ? 


3.  Simplify  Ax*  +  Dx  +  Ey  +  F=Ql)y  translating  the  axes  (a)  if  E  -^  0, 
(b)  if  E  =  0,  and  find  the  point  to  which  the  origin  is  moved. 

A™,  (a)  A*  +  Ey  =  0,  (-  A  , 


*  In  describing  the  final  form  of  the  equation  it  is  unnecessary  to  indicate  by  primes 
what  terms  are  different  from  those  in  (1). 


CONIC  SECTIONS  173 

4.    To  what  types  do  the  loci  of  the  following  equations  belong  ? 

(a)  4z2+?/2-  13x  +  7y  -  1  =  0.  (e)  x2  +  7  y2  -  Sx  +  1  =  0. 

(b)  y2  +  3x  -  4  y  +  9  =  0.  _^C  (f)  x2  +  y2  -  6x  +  8 ?/  =  0. 

(c)  121  x2  -  44  y2  +  68  x  -  4  =  0.       ij y(g)  3x2-4?/2-6y  +  9  =  0. 

(d)  x2+4y-3  =  0.  !  '(h)  x2-8x  +  9?/-  11  =0. 

(i)  The  equations  in  problem  1,  p.  148,  wfiich  do  not  contain  the  xy-term. 


76.  Construction  of  the  locus  of  an  equation  of  the  second 
degree.    To  remove  the  ajy-term  from 


(1  )  Ax2 

it  is  necessary  to  rotate  the  axes  through  an  angle  B  such  that 
(Theorem  VI,  p.  145) 

(2)  tan  2  0  =      B 


A  -C 


while  in  the  formulas  for  rotating  the  axes  [(II),  p.  138]  we  need 
sin  0  and  cos  0.    By  1  and  3,  p.  12,  we  have 


(3)  cos  2  6  =  ± 


+  taii220 


From  (2)  we  can  choose  2  0  in  the  first  or  second  quadrant  so 
the  sign  in  (3)  must  be  the  same  as  in  (2).  6  will  then  be  acute ; 
and  from  15,  p.  13,  we  have 


1  -  cos  2  0  I  +  cos  2  B 

(4) 

In  simplifying  a  numerical  equation  of  the  form  (1)  the  com- 
putation is  simplified,  if  A  =  J52  —  4  A  C  =£  0,  by  first  removing 
the  terms  in  x  and  y  (Theorem  VII,  p.  146)  and  then  the  xy-term. 

Hence  we  have  the 

Rule  to  construct  the  locus  of  a  numerical  equation  of  the  second 
degree. 

First  step.    Compute  A  =  B2  —  4  A  C. 
Second  step.    Simplify  the  equation  by 

(a)  translating  and  then  rotating  the  axes  if  A  =£  0; 

(b)  rotating  and  then  translating  the  axes  if  A  =  0. 


174  ANALYTIC  GEOMETRY 

Third  step.  Determine  the  nature  of  the  locus  by  inspection  of 
the  equation  (§  75,  p.  170). 

Fourth  step.    Plot  all  of  the  axes  used  and  the  locus. 

In  the  second  step  the  equations  for  rotating  the  axes  are 
found  from  equations  (2),  (3),  (4),  and  (II),  p.  138.  But  if  the 
2-y-term  is  lacking,  it  is  not  necessary  to  rotate  the  axes.  The 
equations  for  translating  the  axes  are  found  by  the  Rule  on 
p.  141. 

Ex.  1.    Construct  and  discuss  the  locus  of 

4y2  +  12x-6y  =  0. 


Solution.    First  step.    Here  A  =  42  -4-1-4  =  0. 

Second  step.    Hence  we  rotate  the  axes  through  an  angle  0  such  that, 

by  (2),  _fiU\ 

tan2^T^H-U-<: 

Then  by  (3),  cos  20  =  -!, 

2  1 

and  by  (4),  sin0  =  —  and  cos0  =  -- 

V5  V5 

The  equations  for  rotating  the  axes  [(II),  p.  138]  become 


Substituting  in  the  given  equation,*  we  obtain 

Z'2-Ay'^O. 

V5 

It  is  not  necessary  to  translate  the  axes. 
Third  step.    This  equation  may  be  written 


q 

Hence  the  locus  is  a  parabola  for  which  p  =  —  -,  and  whose  focus  is  on 
the  F'-axis.  Vs 

*When  A=0  the  terms  of  the  second  degree  form  a  perfect  square.    The  work  of 
substitution  is  simplified  if  the  given  equation  is  first  written  in  the  form 


It  can  be  shown  that  when  A  =  0  the  locus  is  always  of  the  parabolic  type. 


CONIC  SECTIONS 


175 


Fourth  step.    The  figure  shows  both  sets  of  axes,*  the  parabola,  its  focus 
and  directrix. 

In  the  new  coordinates  the  focus 

is  the  point  (0,        _)  and  the  direc- 
V      2V5y 


A^s 


trix  is  the  line  y'  = —  (Theorem 

2V5 

IV,  p.  155).  The  old  coordinates  of 
the  focus  may  be  found  by  substi- 
tuting the  new  coordinates  for  x/ 
and  y'  in  (1),  and  the  equation  of 
the  directrix  in  the  old  coordinates 
may  be  found  by  solving  (1)  for  y'  'D 

and  substituting  in  the  equation  given  above. 

Ex.  2.    Construct  the  locus  of 

5  x2  +  6  xy  +  5  y2  +  22  x  —  6  y  +  21  =  0. 
Solution.    First  step.    A-  62  -4-5-  5^0. 

Second  step.    Hence  we  translate  the  axes  first.    It  is  found  that  the  equa- 
tions for  translating  the  axes  are 


and  that  the  transformed  equation  is 


From  (2)  it  is  seen  that  the  axes  must  be  rotated  through  —  •     Hence  we 

set 


yi\ 


"  + 


and  the  final  equation  is 


Third  step.    The  simplified  equa- 
tion may  be  written 


Hence  the  locus  is  an  ellipse  whose  major  axis  is  8,  whose  minor  axis  is  4, 
and  whose  foci  are  on  the  Y"-axis. 

Fourth  step.    The  figure  shows  the  three  sets  of  axes  and  the  ellipse. 


*  The  inclination  of  OX'  is  0,  and  hence  its  slope,  tan  9,  may  be  obtained  from  (4).    In 

is  example  tan0= = — --• — '• 

cos0      V5      > 

method  given  in  the  footnote,  p.  28. 


this  example  tan0= = — =-. 7^  =  2,  and  the  Jf'-axis  may  be  constructed  by  the 

cosfl      V5      V5 


176  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  Simplify  the  following  equations  and  construct  their  loci,  foci,  and 
directrices. 

(a)  3z2  -  4zy  +  8z  -  1  =  0.  Ana.   z"2  -  4y"2j-  1  =  0. 

(b)  4z2  +  4zy  +  y2  +  8z-16y  =  0.         4ns.    5z'2  -  8  V5y'  =  0. 

(c)  41  z2  -  24  zy  +  34  y2  +  25  =  0.  Ans.   x/2  +  2  y"2  +  1  =  0. 
(d) 


(e)  y2  +  6z-6y  +  21  =  0.  -4na.   y'2  -f  6z'  =  0. 

(f)  z2-6zy  +  9y2  +  4z-  12y  +  4  =  0.  Ana.   y"2  =  0. 

(g)  12  zy  -  5y2  +  48  y  -  36  =  0.  Ans.   4  z"2  -  9  y"  =  36. 
(h)  4z2  -I2xy  +  9y2  +  2x-  3y-12  =  0. 

^4ns.    52  y//2  -  49  =  0. 
(i)  14z2-4zy  +  ll?/2-88z  +  34  y  +  149  =  0. 

-4ns.   2  z//2  +  3  y//2  =  0. 
(j)  12z2  +  8zy  +  18?/2  +  48z  +  16y  +  43  =  0. 

Ans.   4z2  +  2y2  =  1. 
(k)  9z2  +  24zy  +  16y2  -36z-48y  +  61  =  0. 

Ana.   z'/2  +  l  =  0. 

(1)  7  z2  +  50  xy  +  7  y2  =  50.  Ana.    16  z/2  -  9  y/2  =  25. 

(m)  z2  +  3  xy  -  3  y2  +  6  z  =  0.  ulns.    21  z//2  -  49  y"2  =  72. 

(n)  16z2  -  24  xy  +  9y2  -  60z  -  80y  +  400  =  0. 

Ana.    2/"2-4z"  =  0. 
(o)  95  z2  +  56  xy  -  10z/2  -  56z  +  20y  +  194  =  0. 

Ana.   6z"2-y//2  +  12=:0. 
(p)  5  z2  -  5  xy  -  7  ?/2  -  165  z  +  1320  =  0.  -4ns.    15  z//2  -  11  y"2  -  330  =  0. 

77.  Systems  of  conies.  The  purpose  of  this  section  is  to 
illustrate  by  examples  and  problems  the  relations  between 
conies  and  degenerate  conies  and  between  conies  of  different 
types. 

A  system  of  conies  of  the  same  type  shows  how  the  degenerate 
conies  appear  as  limiting  forms,  while  a  system  of  conies  of  dif- 
ferent types  shows  that  the  parabolic  type  is  intermediate  between 
the  elliptic  and  hyperbolic  types. 

Ex.  1.    Discuss  the  system  of  conies  represented  by  z2  +  4  y2  =  k. 

Solution.  Since  the  coefficients  of  z2  and  y2  have  the  same  sign,  the  locus 
belongs  to  the  elliptic  type  (Corollary,  p.  172).  When  k  is  positive  the  locus 
is  an  ellipse  ;  when  k  =  0  the  locus  is  the  origin,  —  a  degenerate  ellipse  ;  and 
when  k  is  negative  there  is  no  locus. 


CONIC  SECTIONS 


177 


In  the  figure  the  locus  is  plotted  for  k  =  100,  64,  36, 16,  4,  1,  0.  It  is  seen 
that  as  k  approaches  zero  the  ellipses  become  smaller  and  finally  degenerate 
into  a  point.  As  soon  as  k  becomes  negative  there  is  no  locus.  Hence  the 


point  is  a  limiting  case  between  the  cases  when  the  locus  is  an  ellipse  and 
when  there  is  no  locus.  J 

Ex.  2.    Discuss  the  system  of  conies  represented  by  4  x2  —  16  y2  =  k. 
Solution.    Since  the  coefficients  of  x2  and  y2  have  opposite  signs,  the  locus 

3P 


belongs  to  the  hyperbolic  type.  The  hyperbolas  will  all  have  the  same 
asymptotes  (p.  165),  namely,  the  lines  x  ±  2  y  =  0.  The  given  equation  may 
be  written 


4       16 

The  locus  is  an  hyperbola  whose  foci  are  on  the  X-axis  when  k  is  positive  and 


178 


ANALYTIC  GEOMETRY 


on  the  F-axis  when  k  is  negative.     For  k  =  0  the  given  equation  shows  that 
the  locus  is  the  pair  of  asymptotes. 

In  the  figure  the  locus  is  plotted  for  k  =  256,  144,  64, 16,  0,  -  64,  -  256. 
It  is  seen  that  as  k  approaches  zero,  whether  it  is  positive  or  negative,  the 
hyperbolas  become  more  pointed  and  lie  closer  to  the  asymptotes  and  finally 
degenerate  into  the  asymptotes.  Hence  a  pair  of  intersecting  lines  is  a  lim- 
iting case  between  the  cases  when  the  hyperbolas  have  their  foci  on  the  X-axis 
and  on  the  F-axis. 

Ex.  3.    Discuss  the  system  of  conies  represented  by  y2  =  2  kx  +  16. 

Solution.  As  only  one  term  of  the  second  degree  is  present,  the  locus 
belongs  to  the  parabolic  type  (Corollary,  p.  172).  The  given  equation  may  be 

simplified  (Rule,  p.  141)  by  translating  the  axes  to  the  new  origin  (  — ,  0  Y 
We  thus  obtain  \     k      * 


g 
The  locus  is  therefore  a  parabola  whose  vertex  is  ( »  0)  and  for  which 

p  =  k.    It  will  be  turned  to  the  right  when  k  is  positive,  and  to  the  left  when 
k  is  negative.     But  if  k  =  0,  the  locus  is  the  degenerate  parabola  y  =  ±  4. 


In  the  figure  the  locus  is  plotted  for  k  =  ±  4,  ±  2,  ±  1,  ±  f ,  0.  It  is  seen 
that  as  k  approaches  zero,  whether  it  is  positive  or  negative,  the  vertex  recedes 
from  the  origin  and  the  parabola  lies  closer  to  the  lines  y  =  ±  4  and  finally 
degenerates  into  these  lines.  The  degenerate  parabola  consisting  of  two 
parallel  lines  appears  as  a  limiting  case  between  the  cases  when  the  parab- 
olas are  turned  to  the  right  and  to  the  left. 


CONIC  SECTIONS  179 

PROBLEMS 

1.  Plot  the  following  systems  of  conies. 

(a)  ~  +  |  -  *.       (b)  y*  =  2kx.       (c)  H  -  £  =  *.       (d)  x2  =  2ky  -  6. 

x2      w2 

2.  Plot  the  system  --  (-  —  -t  —  1  for  positive  values  of  k.     What  is  the 

locus  if  k  =  16?    Show  how  the  foci  and  directrices  behave  as  k  increases 
or  decreases  and  approaches  16. 

3.  Plot  the  following  systems  of  conies  and  show  that  all  of  the  conies 
of  each  system  have  the  same  foci. 


4.  Plot  and  discuss  the  system  fcx2  +  2  y2  -  8x  =  0. 

5.  Show  that  all  of  the  conies  of  the  f^  Ovdi-s  -systems  pass  through  the 
points  of  intersection  of  the  conies  obtained  by  setting  the  parentheses  equal 
to  zero.     Plot  the  systems  and  discuss  the  loci  for  the  values  of  k  indicated. 

(a)  (y2-4x)  +  A;(y2  +  4x)  =  0,  k  =  +  1,  -1. 

(b)  (x2  +  y2  -  16)  +  fc(x2  -  ?/2  -  4)  =  0,  k  =  +  1,  -  1,  -  4. 

(c)  (x2  +  y2  -  16)  +  fc(x2  -  y2  -  16)  =  0,  k  =  +  1,  -  1. 


6.  Find  the  equation  of  the  locus  of  a  point  P  if  the  sum  of  its  distances 
from  the  points  (c,  0)  and  (—  c,  0)  is  2  a. 

7.  Find  the  equation  of  the  locus  of  a  point  P  if  the  difference  of  its 
distances  from  the  points  (c,  0)  and  (—  c,  0)  is  2  a. 

8.  Show  that  a  conic  or  degenerate  conic  may  be  found  which  satisfies 
five  conditions,  and  formulate  a  rule  by  which  to  find  its  equation.     Find 
the  equation  of  the  conies 

(a)  Passing  through  (0,  0),  (1,  2),  (1,  -  2),  (4,  4),  (4,  -  4). 

(b)  Passing  through  (0,  0),  (0,  1),  (2,  4),  (0,  4),  (-  1,  -  2). 

The  circle  whose  radius  is  a  and  whose  center  is  the  center  of 
a  central  conic  is  called  the  auxiliary  circle. 

9.  The  ordinates  of  points  on  an  ellipse  and  the  auxiliary  circle  which 
have  the  same  abscissas  are  in  the  ratio  of  6  :  a. 

10.  Show  that  the  locus  of  xy  +  Dx  -f  Ey  +  F  =  0  is  either  an  equilateral 
hyperbola  whose  asymptotes  are  parallel  to  the  coordinate  axes  or  a  pair  of 
perpendicular  lines. 


CHAPTER   IX 
TANGENTS   AND   NORMALS 

78.  The  slope  of  the  tangent.  Let  P^  be  a  fixed  point  on  a 
curve  C  and  let  P2  be  a  second  point  on  C  near  Pj.  Let  P2 
approach  Px  by  moving  along  C.  Then  the  limiting  position 
1-iT  of  the  secant  through  P!  and  P2  is  called  the  tangent  to  C 
at  P!. 

It  is  evHent  that  the  slope  of  P^  is  the  limit  of  the  slope 
of  PiP2.  The  coordinates  of  P2  may  be  written  (xl  +  h,  y±  -f  &), 


T 


where  h  and  k  will  be  positive  or  negative  numbers  according 
to  the  relative  positions  of  P:  and  P2.  The  slope  of  the  secant 
through  P!  and  P2  is  therefore  (Theorem  V,  p.  28) 


(i) 


As  P2  approaches  P!  both  h  and  k  approach  zero,  and  hence  - 
approaches  —  >  which  may  be  any  number  whatever.  The  actual 
value  of  the  limit  of  —  may  be  found  in  any  case  from  the  condi- 
tions that  P!  and  P2  lie  on  C  (Corollary,  p.  46),  as  in  the  example 
following. 

180 


TANGENTS  AND  NORMALS 


181 


Ex.  1.    Find  the  slope  of  the  tangent  to  the  curve  C  :  8  y  =  z8  at  any  point 
Pi  (xi,  ?/i)  on  C. 

Solution.    Let  PI  (xi,  y\)  and  P2  (x\  -f  A,  yi  -f  fc)  be  two  points  on  C. 
Then  (Corollary,  p.  46) 


8  A:  =  Xi8  +  3xi*ft  +  3z 
Subtracting  (2)  from  (3),  we  obtain 


Y 

1  1 

.     /9\ 

i 

i 

Vz/ 

7tj 

—    and         8 

A 

'  / 

or 

IL 

^'/ 

(3)           8 

f 

_J 

—  -« 

•W 

/I 

Subtrac 

f 

'o 

/ 

7 

X 

/ 

/ 

/ 

/ 

/ 

I 

Factori 

y 

* 

/ 

/ 

Factoring,     8  k  =  h  (3  xx2  +  3  xji  +  A2) ; 
k  _  3  x^  +  3  x-Ji  +  A2 
A~  8 

Then,  as  P2  approaches  Pl5  A  and  k  approach 
zero  and  the 


7-    -,    s         i-    •*    . i 
limit  of-=  limit  of  -  —•• 

h  8 

Hence  the  slope  m  of  the  tangent  at  PI  is  m  =  — 

8 


C  is  symmetrical  with  respect  to  0,  and  the  tangents  at  symmetrical  points 
are  parallel  since  only  even  powers  of  Xi  and  yi  occur  in  the  value  of  m.  The 
tangent  at  the  origin  is  remarkable  in  that  it  crosses  the  curve. 

The  method  employed  in  this  example  is  general  and  may  be 
formulated  in  the  following 

Rule  to  determine  the  slope  of  the  tangent  to  a  curve  C  at  a  point 
1\  on  C. 

First  step.  Let  PI  (xlt  y-^)  and  P2  (xl  -f-  ^>  y\  +  &)  b&  tw°  points 
on  C*  Substitute  their  coordinates  in  the  equation  of  C  and 
subtract. 

Second  step.  Solve  the  result  of  the  first  step  for  -^>*  the  slope 
of  the  secant  through  PI  and  P2. 

Third  step.  Find  the  limit  of  the  result  of  the  second  step  when 
h  and  k  approach  zero.  This  limit  is  the  required  slope. 


*  The  solution  will  contain  h  and  k  separately,  so  that  the  equation  is  not  solved  in  the 
ordinary  sense. 


182  ANALYTIC  GEOMETRY 


PROBLEMS 

1.  Find  the  slopes  of  the  tangents  to  the  following  curves  at  the  points 
indicated. 

(a)  2/2  =  8  x,  PI  (2,  4).  Ana.  1. 

(b)  x2  +  j/2  =  25,  Pi  (3,  -  4).  Ans.  f . 

(c)  4x2  +  ?/2  =  16?  Pl  (o,  4).  Ana.  0. 

(d)  x2  -  9  2/2  =  81,  P!  (15,  -  4).  Ana.  -  f. 

2.  Find  the  slopes  of  the  tangents  to  the  following  curves  at  the  point 

(a)  2/2  = 

(b)  162/  =  x4.  -^          A-  ^4ns.  —  • 

O)  4 

3*1 
/rt\     /~2     i     /».2  1A  /A  >i^« 


2/1 
2/1 


79.  Equations  of  tangent  and  normal.    We  have  at  once  the 
Rule  to  find  the  equation  of  the  tangent  to  a  curve  C  at  a  point 

PI  (a>i,  2/i)  on  C. 

First  step.  Find  the  slope  m  of  the  tangent  to  C  at  Pl  (Rule, 
p.  181). 

Second  step.  Substitute  %i,  y^  and  m  in  the  point-slope  form  of 
the  equation  of  a  straight  line  [(V),  p.  86]. 

Third  step.  Simplify  that  equation  by  means  of  the  condition 
that  Pl  lies  on  C  (Corollary,  p.  46). 

Ex.  1.    Find  the  equation  of  the  tangent  to  C  :  8  y  =  x*  at  PI  (xi,  ?/i). 

3xi2 
Solution.    First  step.    From  Ex.  1,  p.  481,  the  slope  is  m  =  —  -  — 

Second  step.    Hence  the  equation  of  the  tangent  is 

3X!\ 

y  -  yi  =  -  -  (*  -  «i), 

or 

(1)  3  Xi2x  -  8  y  -  3  xi3  +  8  yi  =  0. 
Third  step.    Since  PI  lies  on  C,  8  yi  =  Xi3. 
Substituting  in  (1),  we  obtain 

(2)  3xi2x-87/  -2xi3  =  0. 


TANGENTS  AND  NORMALS  183 

The  normal  to  a  curve  C  at  a  point  Pl  on  C  is  the  line  through 
Pl  perpendicular  to  the  tangent  to  C  at  Plt  Its  equation  is  found 
from  that  of  the  tangent  by  the  Rule  on  p.  105,  using  Theorem 
XII,  p.  108. 

Ex.  2.    Find  the  equation  of  the  normal  at  PI  to  the  curve  in  Ex.  1. 

Solution.    The  equation  of  any  line  perpendicular  to  (2)  has  the  form 
(Theorem  XII,  p.  108) 
(3)  8x  +  3xi2y  +  fc  =  0. 

If  PI  lies  on  this  line,  then  (Corollary,  p.  46) 


whence  k  =  —  8  Xi  —  3  x^y^. 

Substituting  in  (3),  the  equation  of  the  normal  is 

8  x  +  3  Xi2y  —  8  Xi  —  3  Xi'2yi  =  0. 

PROBLEMS 

1 .  Find  the  equations  of  the  tangents  and  normals  at  PI  (xi,  y\)  to  the 
curves  in  problem  2,  p.  182. 


Ans.    (a)  yiy  =  3(x  +«i),  y\x  +  3  y  =  x$j\  +  3  yi. 

(b)  Xi3x  -  4  y  =  12  ylf  4  x  +  Xi3y  =  4  xx  + 

(c)  Xix  +  yiy  =  16,  yxx  -  «iy  =  0. 

(d)  XiX  -  yiy  =  4,  y&  +  «iy  = 
(e) 


2.  Find  the  equations  of  the  tangents  and  normals  to  the  following  curves 
at  the  points  indicated. 

(a)  y2  -  8  x  +  4  y  =  0,  (0,  0).  <4ns.  2x-y  =  0,  x  +  2y  =  0. 

(b)  xy  =  4,  (2,  2).  .4ns.  x  +  y  =  4,  x-y  =  0. 

(c)  x2  —  4y2  =  25,  PI(XI,  2/1).  ^Ins.  XiX-4yiy=25,  4y1x+Xiy=5x1yi. 

(d)  x2  +  2xy  =  4,  PI(XI,  yi). 

^Ins.    (xi  +  yi)  x  +  Xiy  =  4,  xxx  -  (xi  +  yjy  =  xi2  -  Xiyi  -  y'i2. 

(e)  y2  = 

(f)  62x2 

-4n«.   62xix  -  a^ii/  =  a262,  a2yix  -f  ft^iy  =  (a2 

(g)  x2  -  y2  +  x3  =  0,  (0,  0).  Ans.    y  =  ±  x,  x  =  =F  y. 


184 


ANALYTIC  GEOMETRY 


80.  Equations  of  tangents  and  normals  to  the  conic  sections. 
Theorem  I.    The  equation  of  the  tangent  to  the  circle 


at  the  point  PI  (xl}  y-^)  on  C  is 

(I)  oc&  +  yjj  —  r*. 

Proof.    Let  PI  (xi,  yi)  and  P2  (xi  +  A,  2/1  +  k)  be  two  points  on  the  circle  C. 

Then  (Corollary,  p.  46) 


+  (yi  +  fc)2  =  r2, 


Subtracting  (1)  from  (2),  we  have 


Y' 


Transposing  and  factoring,  this  becomes 
whence 


Ji          2  yi  +  k 
is  the  slope  of  the  secant  through  PI  and  P2. 

Letting  P2  approach  PI,  h  and  k  approach  zero,  so  that  m,  the  slope  of  the 
tangent  at  P1?  is 

,.    ..    .      2xi  +  h          Xi 
m  =  limit  of  — 


or 


The  equation  of  the  tangent  at  PI  is  then  (Theorem  V,  p.  86) 

Xi 

y  -  2/1  - — -(*.—  £1), 
xix  +  y\y  =  %i2  +  yi2* 

But  by  (1),  X!2  +  yi2  =  r2, 

so  that  the  required  equation  is 

*i*  +  y\y  -  r2.  Q.E.D. 

In  like  manner  we  may  prove  the  following  theorems. 

Theorem  II.    The  equation  of  the  tangent  at  Pl  (xl}  y^)  to  the 
ellipse          b2x2  +  a2y2  =  a2b2    is  b2x^  +  a2yty  =  azbz ; 
hyperbola    b2x2  —  a2y2  =  a2b2    is  I^X^K  —  (tfy^y  =  azbz ; 
parabola  y2  =  2px  is  y±y  =  p(oc  + 


TANGENTS  AND  NORMALS  185 

Theorem  III.    The  equation  of  the  tangent  to  the  locus  of 

Ax2  +  Bxy  +  Cif  +  Dx  +  Ey  +  F  =  0 
the  point  PI(X\,  yi)  on  the  locus  is 


Theorem  III  may  be  stated  in  the  form  of  the 
Rule  to  write  the  equation  of  the  tangent  at  P\(x^  y-^  to  the 
locus  of  an  equation  of  the  second  degree. 

First  step.    Substitute  x^x  and  y^y  for  x2  and  y*,  —  —  -  —  —  for 

xy,  and  -     —  and  for  x  and  y  in  the  given  equation. 

2  Z 

Second  step.    Substitute  the  numerical  values  ofxl  and  ylt  if  given, 
in  the  result  of  the  first  step.      The  result  is  the  required  equation. 
From  Theorem  II,  by  the  method  on  p.  183,  we  obtain 
Theorem  IV.    The  equation  of  the  normal  at  PI  (x^  y^)  to  the 
ellipse          bzx2  +  a2if  =  aW  is  a?y&  -  b^y  =  (a2  -  ft1) 
hyperbola    bV  —  a2?/2  =  a*b2  is  a*y&  +  bzx±y  =  (a2  +  62) 
parabola  y2  =  2px  is  yjc  +  py  =  oc^y^ 

PROBLEMS 

1.  Find  the  equations  of  the  tangents  and  normals  to  the  following  conies 
at  the  points  indicated. 

(a)  3x2  _  10  y2  =  17,  (3,  i).  (C)  2x2  -  y*  =  14,  (3,  -  2). 

(b)  7/2  =  4x,  (9,  -  6).  (d)  x2  +  5y2  =  14,  (3,  1). 

(e)  x2-xy  +  2z-7  =  0,  (3,  2). 

(f)  xy-y2  +  6x  +  Sy  -6  =  0,  (-1,4). 

The  directed  lengths  on  the  tangent  and  normal  from  the  point 
of  contact  to  the  ^Y-axis  are  called  the  length  of  the  tangent  and  the 
length  of  the  normal  respectively.  Their  projections  on  the  X-axis 
are  known  as  the  subtangent  and  subnormal. 

2.  Find  the  subtangents  and  subnormals  in  (a),  (b),  (c),  and  (d),  prob- 
lem 1.  Ana.    (a)  -  ^  ft  ;  (b)  -  18,  2  ;  (o)  -  f  ,  6  ;  (d)  f  ,  -  f  . 

3.  Find  the  lengths  of  the  tangents  and  normals  in  (a),  (b),  (c),  and  (d), 
problem  1.  Ans.    (a)  \  VT81,  T^  VTsT  ;  (b)  6  VlO,  2  VlO  ; 

(c)  §  VlO,  2  VlO  ;  (d)  i  V34,  £  V§4. 


186 


ANALYTIC  GEOMETRY 


4.  Find  the  subtangents  and  subnormals  of  (a)  the  ellipse,  (b)  the  hyper- 

bola, (c)  the  parabola. 

62 

(c)  -2 


-4ns. 


a2  - 


a* 


5.  Show  how  to  draw  the  tangent  to  a  parabola  by  means  of  the  sub- 
normal or  subtangent. 

6.  Prove  that  a  point  PI  on  a  parabola  and  the  intersections  of  the 
tangent  and  normal  to  the  parabola  at  P!  with  the  axis  are  equally  distant 
from  the  focus. 

7.  Show  how  to  draw  a  tangent  to  a  parabola  by  means  of  problem  6. 

8.  The  normal  to  a  circle  passes  through  the  center. 

9.  If  the  normal  to  an  ellipse  passes  through  the  center,  the  ellipse  is  a 
circle. 

10.  The  distance  from  a  tangent  to  a  parabola  to  the  focus  is  half  the 
length  of  the  normal  drawn  at  the  point  of  contact. 

11.  Find  the  equation  of  the  tangent  at  a  vertex  to  (a)  the  parabola; 
(b)  the  ellipse;  (c)  the  hyperbola. 

12.  Find  the  subnormal  of  a  point  PI  on  an  equilateral  hyperbola. 

-4ns.   x\. 

13.  In  an  equilateral  hyperbola  the  length  of  the  normal  at  P!  is  equal  to 
the  distance  from  the  origin  to  PI. 

81.  Tangents  to  a  curve  from  a  point  not  on  the  curve. 

Ex.  1.    Find  the  equations  of  the  tangents  to  the  parabola  y2  =  4  x  which 
pass  through  P2  (—  3,  -  2).  . 

Solution.  Let  the  point  of 
contact  of  a  line  drawn  through 
P2  tangent  to  the  parabola  be 
PI.  Then  by  Theorem  III  the 
equation  of  that  line  is 

Since  P2  lies  on  this  line 
(Corollary,  p.  46), 

(2)  -  2  T/!  =  -  6  +  2  Xi ; 
and  since  PI  lies  on  the  parabola, 

(3)  ?/i2  =  4  xi. 

The  coordinates  of  PI,  the 
point  of  contact,  must  satisfy 
(2)  and  (3).  Solving  them,  we 
find  that  PI  may  be  either  of  the  points  (1,  2)  or  (9,  -  6). 


TANGENTS  AND  NORMALS  187 

If  (1,  2)  be  the  point  of  contact,  the  tangent  line  is,  from  (1), 


or  x  —  y  +  1  =  0. 

If  (9,  -  6)  be  the  point  of  contact,  the  tangent  line  is 

-6y  =  2x  +  18, 
or  x-3y  +  9  =  0. 

The  method  employed  may  be  stated  thus  : 

Rule  to  determine  the  equations  of  the  tangents  to  -a  curve  C 
passing  through  P2(#2,  y^)  not  on  C. 

first  step.  Let  PI  (xi,  y-^  be  the  point  of  tangency  of  one  of  the 
tangents,  and  find  the  equation  of  the  tangent  to  C  at  Pl  (Rule, 
p.  182). 

Second  step.  Write  the  conditions  that  (xz,  yz)  satisfy  the  result 
of  the  first  step  and  (x^  y^)  the  equation  of  C,  and  solve  these  equa- 
tions for  xl  and  y±. 

Third  step.  Substitute  each  pair  of  values  obtained  in  the  second 
step  in  the  result  of  the  first  step.  The  resulting  equations  are  the 
required  equations. 

PROBLEMS 

1.  Find  the  equations  of  the  tangents  to  the  following  curves  which  pass 
through  the  point  indicated  and  construct  the  figure. 

(a)  x2  +  y2  =  25,  (7,  -  1).  Ans.  3x  -  4y  =  25,  4x  +  3y  =  25. 

(b)  y'2  =  4x,  (-  1,  0).  Ans.  y  =  x  +  l,  y  +  x  +  l  =  0. 

(c)  16x2  +  25  y2  =  400,  (3,  -  4).  Ans.  y  +  4  =  0,  3x  -  2y  =  17. 

(d)  8  y  =  x3,  (2,  0).  Ans.  y  =  0,  27  x  -  8  y  -  54  =  0. 

(e)  x2  +  16?/2-100  =  0,  (1,  2).  'Ana.  None. 

(f)  2xy  +  z/2  =  8,  (-  8,  8).        Ans.  2x  +  3y-8  =  0,  4z  +  3y-f8  =  0. 
(g)-  2/2  +  4z-6y  =  0,  (-  *,  -1).  Ans.  2x-3y  =  0,  2x-y  +  2  =  0. 
(h)  x2  +  4  y  =  0,  (0,  -  G).  Ans.  None. 

(i)  x2-3y2  +  2x  +  19  =  0,  (-1,  2). 

Ans.   x-f-3?/-5  =  0,  x-3y  +  7  =  0. 
(j)  y2  =  x^  (.^  o).  Ana.    y  =  0,  3x  -  y  -  4  =  0,  3x  +  y  -  4  =  0. 

2.  Find  the  equations  of  the  lines  joining  the  points  of  contact  of  the 
tangents  in  (a),  (b),  (c),  (f),  (g),  and  (i),  problem  1. 

Ana.   (a)  7x-y  =  25;    (b)  x  =  1  ;    (c)  12x-25y  =  100; 
(f)  x  =  l;    (g)  x-2y  =  0;    (i)  y  =  6. 


188 


ANALYTIC  GEOMETRY 


82.   Properties  of  tangents  and  normals  to  conies. 

Theorem  V.    If  a  point  moves  off  to  infinity  on  the  parabola  y2  =  2px,  the 
tangent  at  that  point  approaches  parallelism 
with  the  X-axis.  '    r 

Proof.    The  equation  of  the  tangent  at  the 
point  PI  (xi,  yi)  is  (Theorem  II,  p.  184) 
y\y  =  px 


Its  slope  is  (Corollary  I,  p.  77) 


Vi 

As  PI  recedes  to  infinity  y±  becomes  infinite, 
and  hence  m  approaches  zero,  that  is,  the  tangent 
approaches  parallelism  with  the  -3T-axis.  Q.  E.  u. 

Theorem  VI.    If  a  point  moves  off  to  infinity  on  the  hyperbola 
62x2  _  a2  ^2  _  a2^ 

the  tangent  at  that  point  approaches  coincidence  with  an  asymptote. 

Proof.    The  equation  of  the  tangent  at  the  point  PI  (xi,  y\}  is  (Theorem 
II,  p.  184) 
(1)  b2XiX  —  a2yiy  =  a262. 

Its  slope  is  (Corollary  I,  p.  77)          m  =  — -• 

d2yi 


As  PI  recedes  to  infinity  x\  and  y\  become  infinite  and  m  has  the  inde- 
terminate form  —  • 

00 

But  since  PI  lies  on  the  hyperbola, 


Dividing  by  a2?/]2,  transposing,  and  extracting  the  square  root, 

^=±>RTi: 

ay  \  ^  2/i2 


Multiplying  by  - , 
a 


TANGENTS  AND  NORMALS 


189 


From  this  form  of  ra  we  see  that  as  y\  becomes  infinite  m  approaches 

±  -i  the  slopes  of  the  asymptotes  [(5),  p.  166],  as  a  limit.    The  intercepts  of 

a       a2  62 

(1)  are  —  and  ---     As  their  limits  are  zero  the  limiting  position  of  the 

tangent  will  pass  through  the  origin.     Hence  the  tangent  at  PI  approaches 
coincidence  with  an  asymptote.  Q.E.D. 

These  theorems  show  an  essential  distinction  between  the  form  of  the 
parabola  and  that  of  the  right-hand  branch  of  the  hyperbola. 

Theorem  VII.     The  tangent  and  normal  to  an  ellipse  bisect  respectively  the 
external  and  internal  angles  formed  by  the  focal  radii  of  the  point  of  contact.* 
Proof.    The  equation  of  the  lines  joining  PI  (zi,  y\)  on  the  ellipse 

&2z2  +  a2?/2  =  a?W 

to  the  focus  F'  (c,  0)  (Theorem  V,  p.  161)  is 
(Theorem  VII,  p.  88) 

yix  +  (c  —  xi)  y  —  cyi  =  0, 
and  the  equation  of  P\F  is 

yix  -  (c  +  zi)  y  +  cyi  =  0. 

The    equation    of    the    tangent    AB   is 
(Theorem  II,  p.  184), 


.A" 


We  shall  show  that  the  angle  0  which  AB  makes  with  P±F'  equals  the 
angle  0  which  P\F  makes  with  AB. 
By  Theorem  ¥.,  p.  100, 


tan  0  - 

~ 


But  since  PI  lies  on  the  ellipse, 


2cyi  -  (a2  -  62) 


and  (Theorem  V,  p.  161) 

22  _ 
Hence 


a2  -  &2  =  c2. 


52 


a'2cyl  - 


cyi  (a 


2  - 


In  like  manner 

tan  <b  •= 


a?cyi  +  (a2  - 
a262  +  &2czi 


*This  theorem  finds  application  in  the  so-called  whispering  galleries. 


190 


ANALYTIC  GEOMETRY 


Hence  tan  6  =  tan  <f> ;  and  since  0  and  0  are  both  less  than  it,  0  =  <f>. 
That  is,  AB  bisects  the  external  angle  of  FPi  and  F'P±,  and  hence,  also, 
CD  bisects  the  internal  angle.  Q.E.D. 

In  like  manner  we  may  prove  the  following  theorems. 

Theorem  VIII.  The  tangent  and  normal  to  an  hyperbola  bisect  respec- 
tively the  internal  and  external  angles  formed  by  the  focal  radii  of  the  point 
of  contact 


Theorem  IX.  The  tangent  and  normal  to  a  parabola  bisect  respectively  the 
internal  and  external  angles  formed  by  the  focal  radius  of  the  point  of  contact 
and  the  line  through  that  point  parallel  to  the  axis.* 

These  theorems  give  rules  for  constructing  the  tangent  and  normal  to  a 
conic  by  means  of  ruler  and  compasses. 

Construction.  To  construct  the  tangent  and  normal  to  an  ellipse  or  hyper- 
bola at  any  point,  join  that  point  to  the  foci  and  bisect  the  angles  formed  by 
these  lines.  To  construct  the  tangent  and  normal  to  a  parabola  at  any  point, 
draw  lines  through  it  to  the  focus  and  parallel  to  the  axis,  and  bisect  the 
angles  formed  by  these  lines. 

The  angle  which  one  curve  makes  with  a  second  is  the  angle  which  the 
tangent  to  the  first  makes  with  the  tangent  to  the  second  if  the  tangents  are 
drawn  at  a  point  of  intersection. 

Theorem  X.    Confocal  ellipses  and  hyperbolas  intersect  at  right  angles. 
Proof.    X,et 


be  an  ellipse  and  hyperbola  with  the  same  foci.     Then 
(3)  o2  -  62  =  a'2  +  6'2. 

For  if  the  foci  are  (±  c,  0),  then  in  the  ellipse  c2  =  a2  -  fc2  and  in  the  hyperbola  c2—  a'2  4-  i'a 
(Theorems  V  and  VI,  p.  161). 


This  theorem  finds  application  in  reflectors  for  lights. 


TANGENTS  AND  NORMALS  191 

The  equations  of  the  tangents  to  (2)  at  a  point  of  intersection  PI  (xi,  y\) 
are  (Rule,  p.  185) 


It  is  to  be  proved  that  the  lines  (4)  are  perpendicular,  that  is  (Corollary 
III,  p.  78),  that 

X!2  tti2 

(5) 


Since  PI  lies  on  both  curves  (2),  we  have 

^  +  ^  =  land^ 
a2       62  a'2 

Subtracting  these  equations,  we  obtain 


But  from  (3),  a2  -  a'2  =  62  +  6'2, 

and  hence  (6)  reduces  to  (5)  and  the  lines  (4)  are  perpendicular.  Q.E.D. 

In  like  manner  we  prove 

Theorem  XI.    Two  parabolas  with  the  same  focus  and  axis  which  are  turned 
in  opposite  directions  intersect  at  right  angles. 

Hence  the  confocal  systems  in  problem  3,  p.  179,  are  such  that  the  two 
curves  of  the  system  through  any  point  intersect  at  right  angles. 

PROBLEMS 

1.  Tangents  to  an  ellipse  and  its  auxiliary  circle  (p.  179)  at  points  with 
the  same  abscissa  intersect  on  the  JT-axis. 

2.  The  point  of  contact  of  a  tangent  to  an  hyperbola  is  midway  between 
the  points  in  which  the  tangent  meets  the  asymptotes. 

3.  The  foot  of  the  perpendicular  from  the  focus  of  a  parabola  to  a  tan- 
gent lies  on  the  tangent  at  the  vertex. 

4.  The  foot  of  the  perpendicular  from  a  focus  of  a  central  conic  to  a  tan- 
gent lies  on  the  auxiliary  circle  (p.  179). 

5.  Tangents  to  a  parabola  from  a  point  on  the  directrix  are  perpendicular 
to  each  other. 

6.  Tangents  to  a  parabola  at  the  extremities  of  a  chord  which  pass  through 
the  focus  are  perpendicular  to  each  other. 

7.  The  ordinate  of  the  point  of  intersection  of  the  directrix  of  a  parabola 
and  the  line  through  the  focus  perpendicular  to  a  tangent  is  the  same  as  that 
of  the  point  of  contact. 


192  ANALYTIC  GEOMETRY 

8.  How  may  problem  7  be  used  to  draw  a  tangent  to  a  parabola  ? 

9.  The  line  drawn  perpendicular  to  a  tangent  to  a  central  conic  from  a 
focus  and  the  line  passing  through  the  center  and  the  point  of  contact  inter- 
sect on  the  corresponding  directrix. 

10.  The  angle  which  one  tangent  to  a  parabola  makes  with  a  second  is 
half  the  angle  which  the  focal  radius  drawn  to  the  point  of  contact  of  the 
first  makes  with  that  drawn  to  the  point  of  contact  of  the  second. 

11.  The  product  of  the  distances  from  a  tangent  to  a  central  conic  to  the 
foci  is  constant. 

12.  Tangents  to  any  conic  at  the  ends  of  the  latus  rectum  (double  chord 
through  the  focus  perpendicular  to  the  principal  axis)  pass  through  the 
intersection  of  the  directrix  and  principal  axis. 

13.  Tangents  to  a  parabola  at  the  extremities  of  the  latus  rectum  are 
perpendicular. 

14.  The  equation  of  the  parabola  referred  to  the  tangents  in  problem  13  is 

x2  -  2 xy  +  y2  -  2V%p (x  +  y)  +  2p2  =  0, 
or  (compare  p.  10)  x^  +  y^  =   v_p  V2. 

15.  The  area  of  the  triangle  formed  by  a  tangent  to  an  hyperbola  and 
the  asymptotes  is  constant. 

16.  The  area  of  the  parallelogram  formed  by  the  asymptotes  of  an 
hyperbola  and  lines  drawn  through  a  point  on  the  hyperbola  parallel  to 
the  asymptotes  is  constant. 

17.  Find  the  length  of  the  tangent  to  a  parabola  at  an  extremity  of  the 
latus  rectum  and  restate  the  equation  of  the  parabola  in  problem  14  in  terms 
of  this  length. 

83.  Tangent  in  terms  of  its  slope.  The  coordinates  of  the 
points  of  intersection  of  a  line  and  conic  are  found  by  solving 
their  equations  (Rule,  p.  69),  which  are  of  the  first  and  second 
degrees  respectively.  To  solve  their  equations  we  eliminate  x 
or  y*  as  may  be  more  convenient,  and  thus  obtain  an  equation 
of  one  of  the  forms 
(1)  Ay*  +  Bij+C  =  0,  Ax2  +  Bx+C  =  0. 

If  the  discriminant  A  =  B2  —  4  A  C  is  zero,  the  roots  of  (1) 
are  real  and  equal  (Theorem  II,  p.  3),  and  hence  the  points  of 
intersection  of  the  line  and  conic  will  coincide,  that  is,  the  line  is 

*  If  one  variable  is  lacking  in  either  equation,  we  usually  solve  that  equation  for  the 
other  variable.  But  for  our  purposes  we  always  eliminate  the  variable  which  occurs  in 
both  equations. 


TANGENTS  AND  NORMALS 


193 


tangent  to  the  conic.  The  equation  obtained  by  setting  the  dis- 
criminant equal  to  zero  is  called  the  condition  for  tangency. 
Hence  the  condition  for  tangency  of  a  line  and  conic  is  found 
by  eliminating  either  x  or  y  from  their  equations  and  setting 
the  resulting  quadratic  equal  to  zero. 

Ex.  1.    Find  the  condition  for  tangency  of  the  line  -  +  -  =  1  and  the  parab- 
ola y2  =  2  px. 

Solution.    Eliminating  x  by  solving  the  first  equation  for  x  and  substituting 
in  the  second,  we  get 

by2  +  2  apy  -  2  abp  =  0. 

The  discriminant  of  this  quadratic  is 

A  =  (2  ap)2  -  4  b  ( -  2  abp)  =  4  op  (op  +  2  62). 
Hence  the  condition  for  tangency  is 

4  ap  (ap  +  2  &2)  =  0  or  ap  (ap  +  2  62)  =  0. 

Ex.  2.    Find  the  equations  of  the  lines  with  the  slope  $  which  are  tangent 
to  the  hyperbola  x2  —  6  y2  +  12  y  — 18  =  0  and  find  the  points  of  tangency. 
Solution.    The  lines  of  the  system 

(1)  V  = 

have  the  slope  $  (Theorem  I,  p.  51). 

Y± 


Solving  (1)  for  x  and  substituting  in  the  given  equation, 

(2)  y2  +  (4fc  -  6)y  +  9  -  2  A:2  =  0. 
Hence  the  condition  for  tangency  is 

(4  k  -  G)2  -  4  (9  -  2  fc2)  =  0. 
Solving  this  equation,  k  =  0  or  2. 
Substituting  in  (1),  we  get  the  required  equations,  namely, 

(3)  x-2y  =  0,  x-2y  +  4  =  0. 

To  find  the  points  of  tangency  we  substitute  each  value  of  k  in  (2),  which 
then  assumes  the  second  form  of  (7),  p.  4,  namely, 

if  k  =  0,  (2)  becomes  (y  -  3)2  =  0  ;  .-.  y  =  3 ; 
if  k  =  2,  (2)  becomes  (y  +  I)2  =  0 ;   .:y  =  -  1. 


194  ANALYTIC  GEOMETRY 

Hence  3  and  —  1  are  the  ordinates  of  the  points  of  contact.    Then,  from  (1), 
if  k  =  0  and  y  =  3,  we  have  x  =  6  ; 
if  k  =  2  and  y  =  —  1,  we  have  x  =  —  6. 

Hence,  if  k  =  0,  the  point  of  contact  is  (6,  3)  ; 

if  k  =  2,  the  point  of  contact  is  (-  6,  -  1). 

The  points  of  contact  may  also  be  found  by  solving  each  of  equations  (3) 
with  the  given  equation. 

The  method  of  obtaining  equations  (3)  may  be  summed  up 
in  the 

Rule  to  find  the  equation  of  a  tangent  in  terms  of  its  slope  m. 

First  step.  Find  the  condition  for  tangency  of  the  line  y  =  mx  -f  k 
and  the  given  conic. 

Second  step.  Solve  the  equation  found  in  the  second  step  for  k 
and  substitute  the  values  found  in  y  =  mx  -f-  k.  The  equations 
obtained  are  those  required. 

By  means  of  this  Rule  we  may  prove 

Theorem  XII.  The  equation  of  a  tangent  in  terms  of  its  slope  m 
to  the 

circle  x2  +  y2  =  r2       is  y  —  mx  ±  r  Vl  +  m2  ; 

ellipse        b2x2  +  a'2y2  =  a2b2    is  y  =  mx  ± 


hyperbola  b2x2  —  a2y2  =  a2b2   is  y  —  mx  ±      «2w2  —  bz  ; 

D 

parabola  y2  =  2px  is  y  =  mx  +  —  -- 


PROBLEMS 

1.  Determine  the  condition  for  tangency  of  the  loci  of  the  following 
equations. 

(a)  4x2  +  y2  -  4  x  -  8  =  0,  y  =  2  x  +  k.  Ans.    k2  +  2  k  -  17  =  0. 

(b)  z2  +  ?/2  =  r2,  4y-3z  =  4A;.  Ans. 


. 

a?       62  a      ft 

.^_?!=i,  £+?=i.  . 

of      6*  a      ft  of      p» 

*In  these  problems  it  is  assumed  that  the  constants  involved  are  not  zero. 


TANGENTS  AND  NORMALS  195 

2.  Find  the  equations  of  the  tangents  to  the  following  conies  which  satisfy 
the  condition  indicated,  and  their  points  of  contact.  Verify  the  latter  approx- 
imately by  constructing  the  figure. 

(a)  y2  =  4  x,  slope  =  \.  Ans.   x-2y  +  4  =  0. 

(b)  x2  +  y2  =  16,  slope  =  —  f  .  Ans.    5x  +  3y±20  =  0. 

(c)  9x2  +  16  y2  =  144,  slope  =  -  \.         Ans.   x  +  4  y  ±  4  VlO  =  0. 

(d)  x2  -  4  y2  =  36,  perpendicular  to6x-4y  +  9  =  0. 

Ans.    2x  +  3y 


(e)  x2  +  2  ?y2  -  x  +  y  =  0,  slope  =  —  1. 

(f)  xy  +  y2  —  4  x  +  8  y  =  0,  parallel  to  2  x  -  4  y  =  7. 

^Ircs.    x  '  =  2  y,  x  -  2  y  +  48  =  0. 

(g)  x2  +  2xy  +  y2  +  8x  -  6y  =  0,  slope  =  f. 

ulns.   4  x  -  3  y  =  0. 

(h)  x2  +  2  xy  —  4  x  +  2  y  =  0,  slope  =  2. 

.    y  =  2  x,  2  x  -  y  +  10  =  0. 


3.  Find  the  equations  of  the  common  tangents  to  the  following  pairs  of 
conies.     Construct  the  figure  in  each  case. 


(a)  7/2  =  5x,  9x2  +  9y2  =16.  Ans.    9 

(b)  9x2  +  162/2  =  144,  7x2-32y2  =  224.  Ans.    ±  x  -  y  ±  5  =  0. 

(c)  x2  +  y2  =  49,  x2  +  2/2  _  20y  +  99  =  0. 

Ans.    ±4x-3y  +  35  =  0,  ±3x  —  4y  +  35  = 


Hint.   Find  the  equations  of  a  tangent  to  each  conic  in  terms  of  its  slope  and  then 
determine  the  slope  so  that  the  two  lines  coincide  (Theorem  III,  p.  79). 

4.  Two  tangents,  one  tangent,  or  no  tangent  can  be  drawn  from  a  point 
•Pi  (xi»  yi)  to  the  locus  of 

(a)  y2  =  2px  according  as  ?/i2  —  1px\  is  positive,  zero,  or  negative. 

(b)  62x2-+  a2y*  =  a262  according  as  fc^2  +  a2yi2  -  a262  is  positive,  zero, 
or  negative. 

(c)  62x2  -  a2y2  =  a262  according  as  62X!2  -  afyi2  -  a?V*  is  negative,  zero, 
or  positive. 

5.  Two  perpendicular  tangents  to 

(a)  a  parabola  intersect  on  the  directrix. 

(b)  an  ellipse  intersect  on  the  circle  x2  +  y2  =  a2  +  fc2. 

(c)  an  hyperbola  intersect  on  the  circle  x2  +  y2  =  a2  —  ft2. 


CHAPTER   X 


CARTESIAN  COORDINATES  IN  SPACE 

84.  Cartesian  coordinates.  The  foundation  of  Plane  Analytic 
Geometry  lies  in  the  possibility  of  determining  a  point  in  the 
plane  by  a  pair  of  real  numbers  (a,  y}  (p.  18).  The  study  of 
Solid  Analytic  Geometry  is  based  on  the  determination  of  a  point 
in  space  by  a  set  of  three  real  numbers  x,  y,  and  2.  This  deter- 
mination is  accomplished  as  follows  : 

Let  there  be  given  three  mutually  perpendicular  planes  inter- 
secting in  the  lines  XX' 9  YY',  and  ZZ1  which  will  also  be  mutually 
perpendicular.  These  three 
planes  are  called  the  coordinate 
planes  and  may  be  distin- 
guished as  the  JCF-plane,  the 
FZ-plane,  and  the  ZX-plane. 
Their  lines  of  intersection 
are  called  the  axes  of  coordi- 
nates, and  the  positive  direc- 
tions on  them  are  indicated 
by  the  arrowheads.*  The 
point  of  intersection  of  the 
coordinate  planes  is  called 
the  origin. 

Let  P  be  any  point  in  space  and  let  three  planes  be  drawn 
through  P  parallel  to  the  co6rdinate  planes  and  cutting  the  axes 
at  A,  B,  and  C.  Then  the  three  numbers  OA  =  x,  OB  =  y,  and 
OC  =  z  are  called  the  rectangular  coordinates  of  P. 

*  XX'  and  ZZ'  are  supposed  to  be  in  the  plane  of  the  paper,  the  positive  direction  on 
XXf  being  to  the  right,  that  on  ZZ'  being  upward.  Y  Y'  is  supposed  to  be  perpendicular 
to  the  plane  of  the  paper,  the  positive  direction  being 4n  front  of  the  paper,  that  is,  from 
the  plane  of  the  paper  toward  the  reader. 

196 


CARTESIAN  COORDINATES  IN  SPACE 


197 


Any  point  P  in  space  determines  three  numbers,  the  coordinates 
of  P.  Conversely,  given  any  three  real  numbers  xy  y,  and  2,  a 
point  P  in  space  may  always  be  constructed  whose  coordinates 
are  x,  y,  and  z.  For  if  we  lay  off  OA  =  x,  OB  =  y,  and  OC  =  z, 
and  draw  planes  through  A,  B,  and  C  parallel  to  the  coordinate 
planes,  they  will  intersect  in  such  a  point  P.  Hence 

Every  point  determines  three  real  numbers,  and  conversely,  three 
real  numbers  determine  a  point. 

The  coordinates  of  P  are  written  (aj,  y,  z),  and  the  symbol 
P(x,  y>  z)  is  to  be  read,  "The  point  P  whose  coordinates  are 
x,  y,  and  «." 

The  coordinate  planes  divide  all  space  into  eight  parts  called 
octants,  designated  by  0-XYZ,  O-X'YZ,  etc.  The  signs  of  the 
coordinates  of  a  point  in  any  octant  may  be  determined  by  the 

Rule  for  signs. 

x  is  positive  or  negative  according  as  P  lies  to  the  right  or  left 
of  the  YZ-plane. 

y  is  positive  or  negative  according  as  P  lies  in  front  or  in  back 

of  the  ZX-plane. 

z  is  positive  or  negative  according  as 
P  lies  above  or  below  the  XY-plane. 

If  the  coordinate  planes   are  not 
mutually  perpendicular,  we  still  have 
an  analogous  system  of  coordinates 
called   oblique   coordinates.      In    this 
system   the  coordinates  of  a   point 
_^  are   its   distances   from   the  coord i- 
x  nate  planes  measured  parallel  to  the 
axes  instead  of  perpendicular  to  the 
planes.     We  shall  confine  ourselves 
to  the  use  of  rectangular  coordinates. 

Points  in  space  may  be  conveniently  plotted  by  marking  the  same  scale  on  XX' 
and  ZZ'  and  a  somewhat  smaller  scale  on  YY'.  Then  to  plot  any  point,  for  example 
(7, 6, 10),  we  lay  off  OA  =  7  on  OX,  draw  AQ  parallel  to  OF  and  equal  to  6  units 
on  OY,  and  QP  parallel  to  OZ  and  equal  to  10  units  on  OZ. 


198  ANALYTIC  GEOMETRY 

PROBLEMS 

1.  What  are  the  coordinates  of  the  origin  ? 

2.  Plot  the  following  sets  of  points. 

(a)  (8,  0,2),  (-3,  4,  7),  (0,0,  5). 

(b)  (4,  -  3,  6),  (-  4,  6,  0),  (0,  8,  0). 

(c)  (10,  3,  -4),  (-4,  0,0),  (0,8,4). 

(d)  (3,  -  4,  -  8),  (-  5,  -  6,  4),  (8,  6,  0). 

(e)  (-4,  -8,  -6),  (3,  0,7),  (6,  -4,2). 

(f)  (-6,  4,  -4),  (0,  -14,C),  (9,  7, -2). 

3.  Where  can  a  point  move  if  x  =  0  ?  if  y  =  0  ?  if  z  =  0  ? 

4.  Where  can  a  point  move  if  x  =  0  and  y  =  0?  if  y  =  0  and  z  =  0? 
if  z  =  0  and  x  =  0  ? 

5.  Show  that  the  points  (x,  y,  z)  and  (—  x,  y,  z)  are  symmetrical  with 
respect  to  the  FZ-plane ;  (x,  y,  z)  and  (x,  —  y,  z)  with  respect  to  the  ZX- 
plane ;   (x,  y,  z)  and  (x,  y,  —  z)  with  respect  to  the  -XT-plane. 

6.  Show  that  the  points  (x,  y,  z)  and  (—  x,  —  y,  z)  are  symmetrical  with 
respect  to  ZZ';  (x,  y,  z)  and  (x,  —  y,  —  2)  with  respect  to  XX'  \  (x,  y,  z)  and 
(—  x,  y,  —  z)  with  respect  to  YY' ;  (x,  y,  z)  and  (—  x,  —  y,  —  z)  with  respect 
to  the  origin. 

7.  What  is  the  value  of  z  if  P  (x,  y,  z)  is  in  the  -XT-plane  ?  of  x  if  P  is  in 
the  YZ-plane  ?  of  y  if  P  is  in  the  Z  JT-plane  ? 

8.  What  are  the  values  of  y  and  z  if  P  (x,  y,  z)  is  on  the  JT-axis  ?  of  z  and 
x  if  P  is  on  the  Y-axis  ?  of  x  and  y  if  P  is  on  the  Z-axis  ? 

9.  A  rectangular  parallelepiped  lies  in  the  octant  O-XYZ  with  three 
faces  in  the  coordinate  planes.     If  its  dimensions  are  a,  6,  and  c,  what  are 
the  coordinates  of  its  vertices  ? 

85.  Orthogonal  projections.  Lengths.  The  definitions  of  the 
orthogonal  projection  (p.  22)  of  a  point  upon  a  line  and  of  a 
directed  length  AB  upon  a  directed  line  hold  when  the  points 
and  lines  lie  in  space  instead  of  in  the  plane.  It  is  evident  that 
the  projection  of  a  point  upon  a  line  may  also  be  regarded  as 
the  point  of  intersection  of  the  line  and  the  plane  passed, 
through  the  point  perpendicular  to  the  line.  As  two  parallel 
planes  are  equidistant,  then  the  projections  of  a  directed  length 
AB  upon  two  parallel  lines  whose  positive  directions  agree  are  equal. 

From  the  preceding  definitions  follows  at  once  as  on  p.  24, 


CARTESIAN  COORDINATES  IN  SPACE 


199 


Theorem  I.    Given  any  two  points 

xz  —  &!  =  projection  of 
7/2  —  i/1  =  projection  of 
zz  —  z±  =  projection  of 


ylt  z^  and  P2(#2,  2/2> 

upon  JOT', 
upon  YY*, 
upon 


Theorem    II.     The    length    I    of   the   line  joining    tivo   points 
(x\,  V\,  «i)   and  P2(x2,  yz,  zz)  is  given  by 


(H) 


/ 

A 

'} 

^t*^^L.  

j* 

B 

D 

/ 
/ 

/     x 

The  proof  is  similar  to  that  for  the  plane, 
p.  24. 

If  we  construct  a  rectangular  parallelepiped 
by  passing  planes  through  PI  and  P2  parallel 
to  the  coordinate  planes,  its  edges  will  be  paral- 
lel to  the  axes  and  equal  numerically  to  the 
projections  of  PiP2  upon  the  axes.  PiP2  will 
be  a  diagonal  of  this  parallelepiped,  and  hence 
t2  will  equal  the  sum  of  the  squares  of  its  three 
dimensions,  that  is,  of  the  numerical  values  of 
x\  —  ic2,  y\  —  y2,  and  zi  —  z2. 


PROBLEMS 

1.  Find  the  length  of  the  line  joining 

(a)  Px(4,  3,  -  2)  to  P2(-  2,  1,  -  5).  Ans.    7. 

(b)  P!  (4,  7,  -  2)  to  P2  (3,  5,  -  4).  Ans.   3. 

(c)  P!(3,  -  8,  6)  to  P2(6,  -  4,  6).  Ans.    5. 

2.  Show  that  the  points  (-  3,  2,  -  7),  (2,  2,  -  3),  and  (-  3,  6,  -  2)  are 
the  vertices  of  an  isosceles  triangle. 

3.  Show  that  the  points  (4,  3,  -  4),  (-2,  9,  -  4),  and  (-2,  3,  2)  are  the 
vertices  of  an  equilateral  triangle. 

4.  Show  that  the  points   (-4,   0,   2),   (-1,   3  Vjj,   2),   (2,  .0,  2),    and 
(-1,  V3,  2+2  V6)  are  the  vertices  of  a  regular  tetraedron. 

5.  What  does  formula  (II)  become  if  PI  and  P2  lie  in  the  XF-plane  ?  in 
a  plane  parallel  to  the  XF-plane  ? 

6.  The  coordinates  (x,  y,  z)  of  the  point  of  division  P  on  the  line  joining 
-Pi  (»i,  yi»  21)  and  P2  (z2,  y2,  z2)   such  that  the  ratio  of  the  segments  is 
p.  p 

:=  X  are  given  by  the  formulas 


PP2 


x  = 


1  +  X 
Hint.   This  is  proved  as  on  p.  32. 


V  = 


y\ 


200  ANALYTIC  GEOMETRY 

7.  The  coordinates  (x,  y,  z)  of  the  middle  point  P  of  the  line  joining 
PI  (a*i,  2/i,  zj)  and  P2(x2,  y2,  22)  are 


8.  Find  the  coordinates  of  the  point  dividing  the  line  joining  the  follow- 
ing points  in  the  ratio  given. 

*  (a)  (3,  4,  2),  (7,  -  6,  4),  X  =  \.  Ans.    (tf,  |,  |). 

(b)  (-  1,  4,  -  6),  (2,  3,  -  7),  X  =  -  3.  Ans.    (f,  f,  -  y). 

(c)  (8,  4,  2),  (3,  9,  6),  X  =  -  I.  Ans.    (V,  f ,  0). 

(d)  (7,  3,  9),  (2,  1,  2),  X  =  4.  Ans.    (3,  |,  y). 

9.  Show  that  the  points  (7,  3,  4),  (1,  0,  6),  and  (4,  5,   -  2)  are  the 
vertices  of  a  right  triangle. 

10.  Show  that  each  of  the  following  sets  of  points  lies  on  a  straight  line, 
and  find  the  ratio  of  the  segments  in  which  the  third  divides  the  line  joining 
the  first  to  the  second. 

(a)  (4,  13,  3),  (3,  6,  4),  and  (2,  -  1,  5).  Ans.    -  2. 

(b)  (4,  -  5,  -  12),  (-  2,  4,  6),  and  (2,  -  2,  -  6).  Ans,    J. 

(c)  (-  3,  4,  2),  (7,  -  2,  6),  and  (2,  1,  4).  Ans.    1. 

11.  Find  the  lengths  of  the  medians  of  the  triangle  whose  vertices  are 
the  points  (3,  4,  -  2),  (7,  0,  8),  and  (-  5,  4,  6).       Ans.  Vll3,  V89,  2  V29. 

12.  Show  that  the  lines  joining  the  middle  points  of  the  opposite  sides  of 
the  quadrilaterals  whose  vertices  are  the  following  points  bisect  each  other. 

(b)  (8,  4,  2),  (0,  2,  5),  (-  3,  2,  4),  and  (8,  0,  -  6). 
(a)  (0,  0,  9),  (2,  6,  8),  (-  8,  0,  4),  and  (0,  -  8,  6). 

(c)  PI(XI,  2/1,  Zi),  P2(x2,  2/2,  z2),  P3  (£3,  2/3,  «s),  P4(«4,  2/4,  z4). 

13.  Find  the  coordinates  of  the  point  of  intersection  of  the  medians  of  the 
triangle  whose  vertices  are  (3,  6,  —  2),  (7,  -  4,  3),  and  (-  1,  4,  -  7). 

Ans.    (3,  2,  -  2). 

14.  Find  the  coordinates  of  the  point  of  intersection  of  the  medians  of 
the  triangle  whose  vertices  are  any  three  points  Pj,  P2,  and  P3. 

Ans.    [i  (xi  +  x2  +  X3),  |  (2/1  +  2/2  +  2/s),  |  (Zi  +  z2  +  z8)']. 


15.  The  three  lines  joining  the  middle  points  of  the  opposite  edges  of  a 
tetraedron  pass  through  the  same  point  and  are  bisected  at  that  point. 

16.  The  four  lines  drawn  from  the  vertices  of  any  tetraedron  to  the  point 
of  intersection  of  the  medians  of  the  opposite  face  meet  in  a  point  which 
is  three  fourths  of  the  distance  from  each  vertex  to  the  opposite  face  (the 
center  of  gravity  of  the  tetraedron). 

17.  The  points  (xi,  2/1,  Zi),  (xi  +  a,  3/i  +  «,  Zi),  and  (xi,  2/1  +  a,  Zi  +  a)  are 
the  vertices  of  an  equilateral  triangle. 


CHAPTER   XI 
SURFACES,  CURVES,  AND  EQUATIONS 

86.  Loci  in  space.  In  Solid  Geometry  it  is  necessary  to  con- 
sider two  kinds  of  loci  : 

1.  The  locus  of  a  point  in  space  which  satisfies  one  given  con- 
dition is,  in  general,  a  surface. 

Thus  the  locus  of  a  point  at  a  given  distance  from  a  fixed  point  is  a  sphere, 
and  the  locus  o'f  a  point  equidistant  from  two  fixed  points  is  the  plane  which  is 
perpendicular  to  the  line  joining  the  given  points  at  its  middle  point. 

2.  The  locus  of  a  point  in  space  which  satisfies  two  conditions  * 
is,  in  general,  a  curve.     For  the  locus  of  a  point  which  satisfies 
either  condition  is  a  surface,  and  hence  the  points  which  satisfy 
both  conditions  lie  on  two  surfaces,  that  is,  on  their  curve  of 
intersection. 

Thus  the  locus  of  a  point  which  is  at  a  given  distance  r  from  a  fixed  point  PI 
and  is  equally  distant  from  two  fixed  points  P%  and  PS  is  the  circle  in  which  the 
sphere  whose  center  is  PI  and  whose  radius  is  r  intersects  the  plane  which  is 
perpendicular  to  PzPs  at  its  middle  point. 


These  two  kinds  of  loci  must  be  carefully  distinguished. 

87.  Equation  of  a  surface.    First  fundamental  problem.   If 

any  point  P  which  lies  on  a  given  surface  be  given  the  coordinates 
(x,  y,  z),  then  the  condition  which  defines  the  surface  as  a  locus 
will  lead  to  an  equation  involving  the  variables  x,  y,  and  «. 

The  equation  of  a  surface  is  an  equation  in  the  variables  x,  y, 
and  z  representing  coordinates  such  that  : 

1.  The  coordinates  of  every  point  on  the  surface  will  satisfy 
the  equation. 

2.  Every  point  whose  coordinates  satisfy  the  equation  will  lie 
upon  the  surface. 

*  The  number  of  conditions  must  be  counted  carefully.  Thus  if  a  point  is  to  be  equi- 
distant from  three  fixed  points  /',,  P2,  and  A,,  it  satisfies  two  conditions,  namely,  of  being 
equidistant  from  /',  and  P2  and  from  P3  and  7'3. 

201 


202  ANALYTIC  GEOMETRY 

If  the  surface  is  defined  as  the  locus  of  a  point  satisfying  one 
condition,  its  equation  may  be  found  in  many  cases  by  a  Rule 
analogous  to  that  on  p.  46. 

Ex.  1.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from 
PI  (3,  0,  -  2)  is  4. 

Solution.  Let  P  (x,  y,  z)  be  any  point  on  the  locus.  The  given  condition 
may  be  written  P  p  _  4 


By  (II),  p.  199,      PXP  =  V(z  -  3)2  +  y*  +  (Z  +  2)2. 


.-.  V(x  -  3)2  +  y2  +  (z  +  2)2  =  4. 
Simplifying,  we  obtain  as  the  required  equation 

z2  +  y2  +  z2  -  6z  +  4  z  -  3  =  0. 

That  ^js  is  the  equation  of  the  locus  should  be  verified  as  In  Ex.  1,  p.  45. 
We  may  easily  prove 
Theorem  I.    The  equation  of  a  plane  which  is 

parallel  to  the  XY-plane  has  the  form  z  =  constant  ; 
parallel  to  the  YZ  -plane  has  the  form  x  =  constant; 
parallel  to  the  ZX-plane  has  the  form  y  =  constant. 

PROBLEMS 

1  .  Find  the  equation  of  the  locus  of  a  point  which  is  (a)  3  units  above  the 
JTF-plane  ;  (b)  4  units  to  the  right  of  the  FZ-plane  ;  (c)  10  units  back  of 
the  ZX-plane.  ,e  *  y\, 

2.  Find  the  equation  of  the  plane  which  is  parallel  to  (a)  the  JTF-plane 
and  4  units  above  it  ;  (b)  the  ZJT-plane  and  3  units  in  front  of  it  ;  (c)  the 
rZ-plane  and  7  units  to  the  left  of  it. 

3.  Find  the  equation  of  the  sphere  whose  center  is  (<x,  /3,  7)  and  whose  radius 
is  r.  Ans.    x2  +  y2  +  z2  -  2  ax  —  2  fiy  —  2  yz  +  a2  +  /32  +  y2  —  r2  =  0. 

4.  What  are  the  equations  of  the  coordinate  planes  ? 

5.  Find  the  equation  of  the  locus  of  a  point  which  is  equally  distant  from 
the  points  (3,  2,  -  1)  and  (4,  -  3,  0).  Ans.    2z  —  10y  +  2z-ll  =  0. 

88.  Equations  of  a  curve.    First  fundamental  problem.   If 

any  point  P  which  lies  on  a  given  curve  be  given  the  coordinates 
(x,  y,  z)}  then  the  two  conditions  which  define  the  curve  as  a  locus 
will  lead  to  two  equations  involving  the  variables  x,  y,  and  z. 


SURFACES,  CURVES,  AND  EQUATIONS 


203 


The  equations  of  a  curve  are  two  equations  in  the  variables 
xy  y,  and  z  representing  coordinates  such  that : 

1.  The  coordinates  of  every  point  on  the  curve  will  satisfy 
both  equations. 

2.  Every  point  whose  coordinates  satisfy  both  equations  will 
lie  on  the  curve. 

If  the  curve  is  defined  as  the  locus  of  a  point  satisfying  two 
conditions,  the  equations  of  the  surfaces  defined  by  each  condi- 
tion separately  may  be  found  in  many  cases  by  a  Kule  analogous 
to  that  on  p.  46.  These  equations  will  be  the  equations  of  the  curve. 

Ex.  1.  Find  the  equations  of  the  locus  of  a  point  whose  distance  from 
the  origin  is  4  and  which  is  equally  distant  from  the  points  PI  (8,  0,  0)  and 
Pa  (0,8,0). 

Solution.    First  step.    Let  P(x,  y,  z) 
be  any  point  on  the  locus. 

Second  step.  The  given  conditions  are 
(1)  PO  =  4,     PPi  =  PP2. 

Third  step.    By  (II),  p.  199, 
PO  =  Vx'2  +  y2  +  z2, 


PPi  =  V  (x  -  8)2  +  3,2  +  za 
PP2  =  Vx2  +  (y  -  8)2  + 
Substituting  in  (1),  we  get 


Vx2  +  y2  +  z2  =  4,     V(x  -  8)2  +  y2  +  z2  =  Vx2  +  (y  -  8)2  +  z2. 
Squaring  and  reducing,  we  have  the  required  equations,  namely, 

x2  +  y2  +  z2  =  16,     x  -  y  =  0. 
These  equations  should  be  verified  as  in  Ex.  1,  p.  45. 

Ex.  2.  Find  the  equations  of  the  circle  lying  in  the  JTr"-plane  whose  center 
is  the  origin  and  whose  radius  is  5. 

Solution.    In  Plane  Geometry  the  equation  of  the  circle  is  (Corollary,  p.  51) 

(2)  x2  +  y2  =  25. 

Regarded  as  a  problem  in  Solid  Geometry  we  must  have  two  equations 
which  the  coordinates  of  any  point  P  (x,  y,  z)  which  lies  on  the  circle  must 
satisfy.  Since  P  lies  in  the  XF-plane, 

(3)  z  =  0. 

Hence  equations  (2)  and  (3)  together  express  that  the  point  P  lies  in  the 
JTF-plane  and  on  the  given  circle.     The  equations  of  the  circle  are  therefore 
x2  +  y2  =  25,     «  =  0. 


204  ANALYTIC  GEOMETRY 

The  reasoning  in  Ex.  2  is  general.     Hence 

If  the  equation  of  a  curve  in  the  XY-plane  is  knoivn,  then  the 
equations  of  that  curve  regarded  as  a  curve  in  space  are  the  given 
equation  and  z  —  0. 

An  analogous  statement  evidently  applies  to  the  equations  of  a 
curve  lying  in  one  of  the  other  coordinate  planes. 
From  Theorem  I,  p.  202,  we  have  at  once 

Theorem  II.    The  equations  of  a  line  which  is  parallel  to 
the  X-axis  have  the  form       y  =  constant ,     z  =  constant; 
the  Y-axis  have  the  form       z  =  constant,     x  =  constant; 
the  Z-axis  have  the  form       x  =  constant,     y  =  constant. 

PROBLEMS 

• 

1.  Find  the  equations  of  the  locus  of  a  point  which  is 

(a)  3  units  above  the  XF-plane  and  4  units  to  the  right  of  the  FZ-plane. 

(b)  5  units  to  the  left  of  the  FZ-plane  and  2  units  in  front  of  the  ZJT-plane. 

(c)  4  units  back  of  the  ZX-plane  and  7  units  to  the  left  of  the  FZ-plane. 

(d)  9  units  below  the  JTF-plane  and  4  units  to  the  right  of  the  FZ-plane. 

2.  Find  the  equations  of  the  straight  line  which  is 

(a)  5  units  above  the  JFF-plane  and  2  units  in  front  of  the  ZJT-plane. 

(b)  2  units  to  the  left  of  the  FZ-plane  and  8  units  below  the  JTF-plane. 

(c)  3  units  to  the  right  of  the  FZ-plane  and  5  units  from  the  Z-axis. 

(d)  13  units  from  the  X-axis  and  5  units  back  of  the  ZJT-plane. 

(e)  parallel  to  the  F-axis  and  passing  through  (3,  7,  —  5). 

(f)  parallel  to  the  Z-axis  and  passing  through  (—4,  7,  6). 

3.  Find  the  equations  of  the  locus  of  a  point  which  is 

(a)  5  units  above  the  XF-plane  and  3  units  from  (3,  7,  1). 

Ans.    z  =  5,  x2  +  y*  +  z2-Qx-Uy-  2z  +  50  =  0. 

(b)  2  units  from  (3,  7,  6)  and  4  units  from  (2,  5,  4). 

Ans.   x2  +  y2  -f  z2  -  6  x  -  14  y  -  K  z  +  90  =  0, 


(c)  5  units  from  the  origin  and  equidistant  from  (3,  7,  2)  and  (-3,  -  7,  -2). 

Ans.   x2  -f  y2  -f  z2  -  25  =  0,  3x  +  7  y  +  2 z  =  0. 

(d)  equidistant  from  (3,  5,  -  4)  and  (-  7,  1,  6),  and  also  from  (4,  —  6,  3) 
and  (-2,  8,  5).  Ans.    5x  +  2y  -  5z  -f-  H^  0,  3x  -  1y  -  z  +  8  =  0. 

(e)  equidistant  from  (2,  3,  7),  (3,  -  4,  6),  and  (4,  3,  -  2). 

Ans.    2x  —  14  ?/  —  2z  +  l  =  0,  x  +y  —  8  z  +  16  =  0. 

» 


SURFACES,  CURVES,  AND  EQUATIONS  205 

4 .  What  are  the  equations  of  the  edges  of  a  rectangular  parallelepiped  whose 
dimensions  are  a,  6,  and  c,  if  three  of  its  faces  coincide  with  the  coordinate 
planes  and  one  vertex  lies  in  O-XYZ  ?  in  O-XY'Z  ?  in  O-X'Y'Z  ? 

5.  What  are  the  equations  of  the  axes  of  coordinates  ? 

6.  The  following  equations  are  the  equations  of  curves  lying  in  one  of  the 
coordinate  planes.     What  are  the  equations  of  the  same  curves  regarded  as 
curves  in  space? 

(a)  y2  =  4  x.  (e)  x2  +  4  z  +  6  x  =  0. 

(b)  x2  +  z2  =  16.  (f )  y2  -  z2  -  4  y  =  0. 

(c)  8x2  -  y2  =  64.  (g)  yz2  -{-  z2  -  6  y  =  0. 

(d)  4z2  +  9?/2  =  36.  (h)  z2-4z2  +  8z  =  0. 

7.  Find  the  equations  of  the  locus  of  a  point  which  is  equally  distant  from 
the  points  (6,  4,  3)  and  (6,  4,  9),  and  also  from  (-  5,  8,  3)  and  (—  5,  0,  3),  and 
determine  the  nature  of  the  locus.  Ans.    z  =  6,  y  =  4. 

8.  Find  the  equations  of  the  locus  of  a  point  which  is  equally  distant  from 
the  points  (3,  7,  —  4),  (—5,  7,  —  4),  and  (—5,  1,  —  4),  and  determine  the 
nature  of  the  locus.  Ans.   x  =  —  1,  y  =  4. 

89.  Locus  of  one  equation.     Second  fundamental  problem. 
The  locus  of  one  equation  in  three  variables  (one  or  two  may  be 
lacking)  representing  coordinates  in  space  is  the  surface  passing 
through  all  points  whose  coordinates  satisfy  that  equation  and 
through  such  points  only. 

The  coordinates  of  points  on  the  surface  may  be  obtained  as  follows : 

Solve  the  equation  for  one  of  the  variables,  say  z,  assume  pairs  of  values  of 

x  and  y,  and  compute  the  corresponding  values  of  z. 

A  rough  model  of  the  surface  might  then  be  constructed  by  taking  a  thin  board 

for  the  -XT-plane,  sticking  needles  into  it  at  the  assumed  points  (x,  y)  whose 

lengths  are  the  computed  values  of  z,  and  stretching  a  sheet  of  rubber  over  their 

extremities. 

The  second  fundamental  problem,  namely,  of  constructing  the 
locus,  is  usually  discarded  in  space  on  account  of  the  mechanical 
difficulties  involved. 

90.  Locus  of  two  equations.     Second  fundamental  problem. 
The  locus  of  two  equations  in  three  variables  representing  coor- 
dinates in  space  is  the  curve  passing  through  all  points  whose 
coordinates  satisfy  both  equations  and  through  such  points  only. 

The  coordinates  of  points  on  the  curve  may  be  obtained  as  follows  : 
Solve  the  equations  for  two  of  the  variables,  say  x  and  y,  in  terms  of  the  third, 
z,  assume  values  for  z,  and  compute  the  corresponding  values  of  x  and  y. 


206 


ANALYTIC  GEOMETRY 


91.  Discussion  of  the  equations  of  a  curve.     Third  funda- 
mental problem.    The  discussion  of  curves  in  Elementary  Ana- 
lytic Geometry  is  largely  confined  to  curves  which  lie  entirely  in 
a  plane  which  is  usually  parallel  to  one  of  the  coordinate  planes. 
/  Such  a  curve  is  defined  as  t.hq  intfp.raaatian  of  a  given  surface 
/with  a  plane  parallel  to  one  of  the  coordinate  planes.    The  method 
\  |  of  determining  its  nature  is  illustrated  in 

Ex.  1.    Determine  the  nature  of  the  curve  in  which  the  plane  z  =  4  inter- 
sects the  surface  whose  equation  is  y2  +  z2  =  4  x. 

Solution.    The  equations  of  the  curve  are,  by  definition, 

(1)  2/2  +  22  =  4*,       z  =  ± 

Eliminate  z  by  substituting  from  the  second  equation  in  the  first.    This  gives 

(2)  y*  _  4  x_+16  =JXL   zj=  4. 
Equations  (2)  are  also  the  equations  of  the  curve. 

For  every  set  of  values  of  (x,  y,  z)  which  satisfy  both  of  equations  (1)  will  evidently  • 
satisfy  both  of  equations  (2),  and  conversely. 


If  we  take  as  axes  in  the  plane  z  =  4  the  lines  O'X'  and  &Y'  in  which  the 
[plane  cuts  the  ZX-  and  FZ-planes,  then  the  equation  of  the  curve  when 
referred  to  these  axes  is  the  first  of  equations  (2),  namely, 
i(3)  y2  -4x  +  10  =  0. 

For  the  second  of  equations  (2)  is  satisfied  by  all  points  in  the  plane  of  X',  0',  and  Y't 
and  the  first  of  equations  (2)  is  satisfied  by  the  points  in  that  plane  lying  on  the  curve  (3), 
because  the  values  of  the  first  two  coordinates  of  a  point  are  evidently  the  same  when 
referred  to  the  axes  O'X',  O'Y',  and  O'Z  a,s  when  referred  to  the  axes  OX,  OY,  and  OZ. 

The  locus  of  (3)  is  a  parabola  (Rule,  p.  173)  whose  vertex,  in  the  plane 
z  =  4,  is  the  point  (4,  0)  for  which  p  =  2. 


SURFACES,  CURVES,  AND  EQUATIONS  207 

The  method  employed  in  Ex.  1  enables  us  to  state  the 

Rule  to  determine  the  nature  of  the  curve  in  which  a  plane  par- 
allel to  one  of  the  coordinate  planes  cuts  a  given  surface. 

First  step.  Eliminate  the  variable  occurring  in  the  equation  of 
the  plane  from  the  equations  of  the  plane  and  surface.  The  result 
is  the  equation  of  the  curve  referred  to  the  lines  in  which  the  given 
plane  cuts  the  other  two  coordinate  planes  as  axes. 

Second  step.  Determine  the  nature  of  the  curve  obtained  in  the 
second  step  by  the  methods  of  Plane  Analytic  Geometry. 

PROBLEMS 

1.  Determine  the  nature  of  the  following  curves  and  construct  their  loci. 

(a)  x2  -  4  2/2  =  8z,  z  =  8.  (e)  z2  +  4  7/2  +  9^2  =  36j  y  =  im 

(b)  x2  +  9y2  =  9z2,  z  =  2.  (f)  z2  -  4  ?/2  +  zi  =  25,  x  =  -  3. 

(c)  z2-4?/2  =  4z,  y  =  -2.  u      (g)  z2-?/2-4z2  +  6z  =  0,  x=2. 

(d)  x2  +  y2  +  z2  =  25,  x  =  3.  (h)  y2  +  z2  -  4z  +  8  =  0,  y  =  4. 

2.  Construct  the  curves  in  which  eich  of  the  following  surfaces  intersect 
the  coordinate  planes. 

(a)  x2  +  4  s/2  +  I6z2  =  64.  (d)  z2  +  9y2  =  102;. 

(b)  x2  +  4  ?/  -  16  z2  =  64.  (e)  x2  -  9  y2  =  10  z. 

(c)  x2  -  4  y2  -  16  z2  =  64.  (f)  «2  +  4  y2  -  16  z2  =  0. 

3.  Determine  the  nature  of  the  intersection  of  the  surface  2  x  +  y  =  2  z 
with  the  plane  y  =  k ;   with  the  plane  z  =  fc'.     How  does  the  intersection 
change  as  k  or  k'  changes  ?  What  idea  of  the  form  of  the  surface  is  obtained  ? 

4.  Determine  the  nature  of  the  intersection  of  the  surface  x2  +  y2  +  4  z2  =  64 
with  the  plane  z  =  k.     How  does  the  curve  change  as  k  increases  from  0 
to  4  ?  from  —  4  to  0  ?     What  idea  of  the  appearance  of  the  surface  is  thus 
obtained  ? 

5.  Determine  the  nature  of  the  intersection  of  the  surface  4x  —  2y  =  4 
with  the  plane  y  =  k ;  with  the  plane  z  =  k'.     How  does  the  intersection 
change  as  k  or  k'  changes  ?     What  idea  of  the  form  of  the  surface  is  obtained  ? 

92.  Discussion  of  the  equation  of  a  surface.  Third  funda- 
mental problem. 

Theorem  III.  The  locus  of  an  algebraic  equation  passes  through 
the  origin  if  there  is  no  constant  term  in  the  equation. 

The  proof  is  analogous  to  that  of  Theorem  VI,  p.  66. 


208 


ANALYTIC  GEOMETRY 


Theorem  IV.  If  the  locus  of  an  equation  is  unaffected  by  chang- 
ing the  sign  of  one  variable  throughout  its  equation,  then  the  locus 
is  symmetrical  with  respect  to  the  coordinate  plane  from  which  that 
variable  is  measured.  'A  "fl"4*  "  *JX  'M  fr^w^ 

If  the  locus  is  unaffected  by  changing  the  signs  of  two  variables 
throughout  its  equation,  it  is  symmetrical  with  respect  to  the  axis 
along  which  the  third  variable  is  measured. 

If  the  locus  is  unaffected  by  changing  the  signs  of  all  three  variables 
throughout  its  equation,  it  is  symmetrical  with  respect  to  the  origin. 

The  proof  is  analogous  to  that  of  Theorem  IV,  p.  65. 

Rule  to  find  the  intercepts  of  a  surface  on  the  axes  of  coordinates. 
Set  each  pair  of  variables  equal  to  zero  and  solve  for  real  values 
of  the  third. 

The  curves  in  which  a  surface  intersects  the  coordinate  planes 
are  called  its  traces  on  th6  coordinate  planes.  From  the  first 
step  of  the  Rule,  p.  207,  it  is  seen  that 

The  equations  of  the  traces  of  a  surface  are  obtained  by  succes- 
sively setting  x  =  0,  y  =  0,  and  z  =  0  in  the  equation  of  the  surface. 

By  these  means  we  can  determine  some  properties  of  the  surface. 
The  general  appearance  of  a  surface  is  determined  by  considering 
the  curves  in  which  it  is  cut  by  a  system  of  planes  parallel  to  each 
of  the  coordinate  planes  (Eule,  p.  207).  This  also  enables  us  to 

determine  whether  the  sur- 


\ 


face  is  closed  or  recedes  to 
infinity. 

Ex.  1.  Discuss  the  locus  of 
the  equation  y2  +  &  =  4  x. 

Solution.  1.  The  surface 
passes  through  the  origin  since 
there  is  no  constant  term  in 
its  equation. 

2.  The  surface  is  sym- 
metrical with  respect  to  the 
.XF-plane,  the  ZX-plane,  and 
the  X-axis. 

For  the  locus  of  the  given  equation  is  unaffected  by  changing  the  sign  of  z,  of  y,  or  of 
both  together. 


SURFACES,  CURVES,  AND  EQUATIONS  209 

3.  It  cuts  the  axes  at  the  origin  only. 

4.  Its  traces  are  respectively  the  point-circle  y2  +  z2  =  0  and  the  parabolas 
Z2  =  4zandy2  =  4z. 

5.  It  intersects  the  plane  x  =  k  in  the  curve  (Rule,  p.  207) 

This  curve  is  a  circle  whose  center  is  the  origin,  that  is,  is  on  the  JT-axis, 
and  whose  radius  is  2  V&  if  k  >  0,  but  there  is  no  locus  if  k  <  0.  Hence  the 
surface  lies  entirely  to  the  right  of  the  FZ-plarie. 

If  k  increases  from  zero  to  infinity,  the  radius  of  the  circle  increases  from 
zero  to  infinity  while  the  plane  z  =  k  recedes  from  the  FZ-plane. 

The  intersection  of  a  plane  z  =  k  or  y=k',  parallel  to  the  XY-  or  ZX-plane, 
is  seen  (Rule,  p.  207)  to  be  a  parabola  whose  equation  is  (compare  Ex.  1,  p.  206) 


These  parabolas  are  found  to  have  the  same  value  of  p,  namely,  p  =  2, 
and  their  vertices  recede  from  the  YZ-  or  ZJT-plane  as  k  or  k'  increases 
numerically. 

PROBLEMS 

1 .  Discuss  the  loci  of  the  following  equations. 

(a)  x2  +  z2  =  4  x.  (f )  x2  +  2/2  -  z2  =  0. 

(b)  z2  +  y2  + 4z2  =  16.  (g)  z2  -  2/2  _  Z2  =  9. 

(c)  z2  +  2/2  _  4z2  =  16.  (h)  z2  +  ?/2  -  z2  +  2xy  =  0. 

(d)  6o;  +  42/4-  3z  =  12.  (i)  x  +  y  -  6z  =  6. 

(e)  3x  -f  2y  +  z  =  12.  (j)  y2  +  z2  =  25. 

2.  Show  that  the  locus  of  Ax  +  By  +  Cz  +  D  =  0  is  a  plane  by  considering 
its  traces  on  the  coordinate  planes  and  the  sections  made  by  a  system  of 
planes  parallel  to  one  of  the  coordinate  planes. 

3.  Find  the  equation  of  the  locus  of  a  point  which  is  equally  distant  from 
the  point  (2,  0,  0)  and  the  FZ-plane  and  discuss  the  locus. 

Ans.    2/2  +  z2-4z  +  4  =  0. 

4.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the 
point  (0,  0,  3)  is  twice  its  distance  from  the  XF-plane  and  discuss  the  locus. 

Ans.   x2  +  y2  -  3  z2  -  6  z  +  9  =  0. 

5.  Find  the  equation  of  the  locus  of  a  point  whose  distance  from  the  point 
(0,  4,  0)  is  three  fifths  its  distance  from  the  ZJT-plane  and  discuss  the  locus. 

Ans.   25x2  +  162/2  +  25 s2  -  200 y  +  400  =  0. 


210  AXALYTIC  GEOMETRY 

Consider  now  in  turn  loci  defined  by  equations  of  the  first  and 
second  degree  in  x}  y,  and  z. 

EQUATIONS   OF   THE   FIRST  DEGREE 

93.  Plane  and  straight  line.   We  confine  ourselves  to  the  two 
theorems. 

Theorem  V.    Plane.     The  locus  of  the  general  equation  of  the 
first  degree  in  x,  y,  and  z, 
(V)  Ax  +-  By  +  Cz  +  D  =  O, 

isM  plane. 
^  The  proof  follows  the  method  explained  in  problem  2,  p.  209. 

Theorem  VI.    Straight  line.    The  locus  of  two  equations  of  the 

first  degree, 

-  C^z  +  D1  =  0, 


is  a  straight  line  unless  the  coefficients  ofx,  y,  and  z  are  proportional. 

The  proof  follows  at  once  from  Theorem  V  and  Elementary  Geometry. 
If  the  coefficients  are  proportional,  it  is  readily  seen  that  the  traces  (p.  208) 
of  the  planes  are  parallel  (p.  78),  and  hence  the  planes  are  parallel. 

To  plot  a  straight  line  we  need  to  know  only  the  coordinates  of  two  points 
on  the  line.  The  easiest  points  to  obtain  are  usually  those  lying  in  the 
coordinate  planes,  which  we  get  by  setting  one  of  the  variables  equal  to  zero 
and  solving  for  the  other  two.  If  a  line  cuts  but  one  of  the  coordinate 
planes,  we  get  only  one  point  in  this  way,  and  to  plot  the  line  we  draw  a  line 
through  that  point  parallel  to  the  axis  which  is  perpendicular  to  that  plane. 

PROBLEMS 

1.  Find  the  intercepts  on  the  axes  and  the  traces  on  the  coordinate  planes 
of  each  of  the  following  planes  and  construct  the  figures. 

(a)  2»  +  3y  +  4z-24  =  0.  (e)  5z-7y-35  =  0. 

(b)  7z-3y  + z-21  =  0.  (f)  4z  +  3z  +  36  =  0. 

(c)  %'-7y-93  +  63  =  0.  (g)  6y  -  83  -  40  =  0. 

(d)  6x  +  4y  -  z  +  12  =  0.  (h)  3z  +  5z  +  45  =  0. 

2.  Find  the  points  in  which  the  following  lines  pierce  the  coordinate  planes 
and  construct  the  lines. 

(a)  2z  +  y-z  =  2,  x-y  +  2z  =  4.         (c)  x  +  2y  =  8,  2x  -  4y  =  7. 

(b)  4z  +  3?/-6z  =12,  4z-3?/  =  2.      (d)  y  +  z  =  4,  x-y  +  2z=10. 


SURFACES,  CURVES,  AND  EQUATIONS  211 

3.  Find   the   equation   of  the  plane  which  passes   through  the  points 
(1,  1,  -1),  (-2,  -2,  2),  and(l,  -1,  2).  Ans.   x-3y-2z  =  Q. 

4.  Find  the  equation  of  the  plane  whose  intercepts  are  a,  6,  c. 

' 


5.  The  system  of  planes  passing  through  the  line 

AIX  +  Biy  +  Ci«  +  DI  =  0,       A2x  +  B^y  +  C2z  +  A  =  0 
is  represented  by 

AIX  +  #iy  +  Ci«  +  A  +  A:  (^I2x  +  B2y  +  C2z  +  J>2)  =  0, 
where  k  is  an  arbitrary  constant. 

6.  Find  the  equation  of  the  plane  determined  by  the  line  2x  +  y  —  4  =  0, 
y  +  2  z  —  0,  and  the  point  (2,  -  1  ,  1). 


EQUATIONS  OF  THE  SECOND  DEGREE 

The  locus  of  an  equation  of  the  second  degree,  of  which  the 
most  general  form  is 

(1)     Axt  +  Bif  +  Cz*  -f  Dyz  +  Ezx  +  Fxy  +  Gx  +  Hy  +  Iz+K  =  0, 

is  called  a^uadric  surface  or  conicoid.  By  methods  similar  to  those 
employed  in  Chapter  VIII,  p.  172,  it  may  be  shown  that  the 
locus  of  (1)  is  a  pair  of  planes,  one  plane,  a  straight  line,*  or  one 
of  the  loci  about  to  be  discussed. 

94.  The  sphere,  f 

Theorem  VII.    The  equation  of  the  sphere  whose  center  is  the 
point  (a,  /?,  y)  and  whose  radius  is  r  is 

(a?-a)2  +  (2/-^)2  +  (^-y)2  =  r2,  or 
(VII)  «2  +  2/2  +  *2  -  2  ax  -  2ptj  -Zyz  +  a*  +^  +  y2  -  r2  =  O. 


Proof.  Let  P  (x,  y,  z)  be  any  point  on  the  sphere,  and  denote 
the  center  of  the  sphere  by  C.  Then,  by  definition,  PC  =  r. 
Substituting  the  value  of  PC  given  by  (II),  p.  199,  and  squar- 
ing, we  obtain  (VII).  Q.E.D. 

*For  example,  the  locus  of  x9  +  7/2  =  0  is  the  Z-axis.  It  is  to  be  regarded  as  a  special 
case  of  a  cylinder  (Theorem  IX,  p.  213). 

t  In  Analytic  Geometry  the  terms  sphere,  cylinder,  and  cone  are  usually  used  to 
denote  the  spherical  surface,  cylindrical  surface,  and  conical  surface  of  Elementary 
Geometry,  and  not  the  solids  bounded  wholly  or  in  part  by  such  surfaces. 


212  ANALYTIC  GEOMETRY 

Theorem  VIII.    The  locus  of  an  equation  of  the  form 
(VIII)  v2  +  y*  +  z*  +  ax  +  ify  +  j[z  +  K  =  o 

is  determined  as  follows  : 

(a)    When  G*  +  H*  +  72  —  4  K  >  0,  the  locus  is  a  sphere  whose 
center  is  (—  J  G,  —  J  H,  —  J  /)  <md  whose  radius  is 

r  = 


JFftm  £2  +  H2  +  72  -  4  ^  =  0,  *Ae  locus  is  the  point-sphere 


(c)    JFAm  <72  +  #2  +  /'  -  4  tf  <  0,  fAere  is  no  locus. 
The  method  of  proof  is  similar  to  that  on  p.  115. 

PROBLEMS 

1.  Find  the  equation  of  the  sphere  whose  center  is  the  point 

(a)  (or,  0,  0)  and  whose  radius  is  a.  Ans.   x2  +  y'2  +  z2  —  2  ax  =  0. 

(b)  (0,  £,  0)  and  whose  radius  is  0.  Ans.   x2  +  y2  +  z2  —  2  £y  =0. 

(c)  (0,  0,  7)  and  whose  radius  is  7.  -4ns.   x2  -f  y2  -f  z2  —  2  72  =0. 

2.  Determine  the  nature  of  the  loci  of  the  following  equations  and  find  the 
center  and  radius  if  the  locus  is  a  sphere,  or  the  coordinates  of  the  point- 
sphere  if  the  locus  is  a  point-sphere. 

(a)  x2  +  y2  +  z2  -  6  x  +  4  z  =  0.         (c)  z2  +  y2  +  z2  +  4  x  -  z  +  7  =  0. 

(b)  x2  +  y2  +  z2  -f  2x  -  4  y  -  5  =  0.  (d)  x2  +  y2  +  z2  -  12x  +  6y  +  4  z  =  0. 

3.  Find  the  equation  of  the  sphere  which 

(a)  has  the  center  (3,  0,  —  2)  and  passes  through  (1,  6,  —  5). 

Ans.    x2  +  y2  +  z2  -  6  x  +  4  z  —  36  =  0. 

(b)  passes  through  the  points  (0,  0,  0),  (0,  2,  0),  (4,  0,  0),  and  (0,  0,  -  6). 

Ans.    x2  +  y2  +  z*  -4x-2y  +  6  z  =  0. 

(c)  has  the  line  joining  (4,  —  6,  5)  and  (2,  0,  2)  as  a  diameter. 

Ans.   x2  +  y2  +  z2  -6z  +  6y-  7z  +  18  =  0. 

4.  Given  two  spheres  Si  :  x2  +  y2  +  z2  +  £i£  +  -H"i2/  +  /i«  +  -K'l  =  0  and 
£2  :  «2  +  2/2  4-  z2  +  #2Z  +  #2^  +  I2z  +  -K"2  =  0  ;  show  that  the  locus  of 

8k  :  x2  +  y2  +  z*  +  GIX  +  H&  +  IiZ  +  KI 

+  fc  (x«  +  y2  +  z2  +  G2z  +  £T2y  +  12«  +  JST2)  =  0 
is  a  sphere  except  when  k  =  —  1.     In  this  case  the  locus  is  a  plane. 

5.  The  center  of  the  sphere  Sk  in  problem  4  lies  on  the  line  of  centers  of  Si 
and  S2  and  divides  it  into  segments  whose  ratio  is  equal  to  k. 

*  That  is,  a  point  or  sphere  of  radius  zero. 


SURFACES,  CURVES,  AND  EQUATIONS 


213 


6.  When  two  spheres  Si  and  S2  (problem  4)  intersect,  the  system  Sk  con- 
sists of  all  spheres  passing  through  their  circle  of  intersection. 

7.  When  the  spheres  Si  and  S2  (problem  4)  are  tangent,  the  system  Sk 
consists  of  all  spheres  tangent  to  Si  and  S2  at  their  point  of  tangency. 

95.  Cylinders. 

Ex.  1.    Determine  the  nature  of  the  locus  of  y2  =  4x. 

Solution.    The  intersection  of  the  surface  with  a  plane  parallel  to  the 
FZ-plane,  x  =  k,  are  the  lines  (Rule,  p.  207) 
(1)     x  =  k,    y  =  ±2  Vfc, 
which  are  parallel  to  the  Z-axis 
(Theorem  II,  p.  204).    If  k  > 0, 
the  locus  of  equations  (1)  is  a 
pair  of  lines ;  if  k  =  0,  it  is  a 
single  line  (the  Z-axis) ;  and 
if  k  <  0,  equations  (1)  have  no 

locus. 
f*^~\ 

Similarly,  the  intersection 
with  a  plane  parallel  to  the 
ZJT-plane,  y  =  k,  is  a  straight 
line  whose  equations  are  (Rule, 
p.  207) 

and  which  is  therefore  parallel  to  the  Z-axis. 

The  intersection  with  a  plane  parallel  to  the  -ZT"-plane  is  the  parabola 

For  different  values  of  k  these  parabolas  are  equal  and  placed  one  above 
another. 

It  is  therefore  evident  that  the  surface  is  a  cylinder  whose  elements  are 
parallel  to  the  Z-axis  and  intersect  the  parabola  in  the  JTF-plane 

y2  =  4  x,     z  =  0. 

It  is  evident  from  Ex.  1  that  the  locus  of  any  equation  which 
contains  but  two  of  the  variables  x,  y,  and  z  will  intersect  planes 
parallel  to  two  of  the  coordinate  planes  in  one  or  more  straight 
lines  parallel  to  one  of  the  axes  and  planes  parallel  to  the  third 
coordinate  plane  in  equal  curves.  Such  a  surface  is  evidently  a 
cylinder^  Hence 

Theorem  IX.  The  locus  of  an  equation  in  which  one  variable  is 
lacking  is  a  cylinder  whose  elements  are  parallel  to  the  axis  along 
which  that  variable  is  measured. 


214 


ANALYTIC  GEOMETRY 


96.  Cones. 

Ex.  1.  Determine  the  nature  of  the  locus 
of  the  equation  16  x2  +  y'2  -  z2  =  0. 

Solution.    Let  PI  (xi,  y^  Zi)  ne  a  point  on 
a  curve  C  in  which  the  locus  ffltersect- 
plane,  for  example,  z  =  k.     Then 

(1)         16xi«  +  y1a-2!1a  =  0,     zl=k. 

The  origin  OTies  on  the  surface  (Theorem 
III,  p.  207).     It  may  be  shown*  that  the  line 
OPi  lies  entirely  on  the  surface,  and  therefore 
that  the  surface  is  a  cone  whose  vertexis  tl> 
origin. 

In  the  same  way  the  locus  may  be 
shown  to  be  a  cone  whenever  the  equa- 
tion of  the  surface  is  homogeneous  f  in 
the  yariables  x,  y,  anjjs.  Hence 

Theorem  X.  The  locus  of  an  equation 
which  is  homogeneous  in  the  variables  x, 
y,  and  z  is  a  cone  whose  vertex  is  the 
origin. 

PROBLEMS 


1  .  Determine  the  nature  of  the  following  loci  ;  discuss  and  construct  them. 

(a)  x2  +  ?/  =  36.  (e)  x2  -  y2  +  36  z2  =  0. 

(b)  x2  +  2/2  =  z2.        -  (f  )  y2  -  16  x2  +  4  z*  =  0. 
(e)  y'2  +  4  z2  =  0. 

(d)  x2  -  z2  =  16. 


(g)  x2  +  16y2  -  4x  =  0. 
(h)'  x2  +  yz  =  0. 


2.  Find  the  equations  of  the  cylinders  whose  directrices  are  the  following 
curves  and  whose  elements  are  parallel  to  one  of  the  axes. 

(a)  y2  +  z2  —  4  y  =  0,  x  =  0.  (c)  &2x2  -  a2^2  =  a262,  z  =  0. 

(b)  z2  +  2  x  =  8,  y  =  0.  (d)  y2  +  2pz  =  0,  x  =  0. 

3.  Discuss  the  following  loci. 

(a)  x2  +  y'2  =  z2  tan2  7.         (c)  z2  +  x2  =  y2  tan2/3.         (e)  y2  +  z2  =  r2. 
<b)  y2  +  z2  =  x2  tan2  a.        (d)  x2  +  y2  =  r2.  (f  )  z2  +  x2  =  r2. 

*  TAe  Elements  of  Analytic  Geometry,  Smith  and  Gale,  p.  385. 

t  An  equation  is  homogeneous  in  x,  y  ,  and  z  when  all  the  terms  in  the  equation  are  of 
the  same  degree  (footnote,  p.  10). 


SURFACES,  CURVES,  AND  EQUATIONS 


215 


97.  Non-degenerate  quadrics.  If  the  locus  of  an  equation  of 
the  second  degree  is  a  cone,  cylinder,  or  pair  of  planes,  it  is  called 
a  degenerate  quadric,  while  the  surfaces  now  to  be  considered  are 
distinguished  as  non-degenerate  quadrics. 

The  lo«  as  of  the  equation 


,,. 


which  contains  only  the  squares  of  the  variables  with  positive 

coefficients,   is    called  an 

ellipsoid.    The  sections  of 

the   surface    formed    by 

planes  parallel  to  any  one 

of  tfye  coordinate  planes 

(Rule,  p.  207)  are  conies 

of  the  elliptic  type.     If 

a  =  b  =  c,  the  locus  of  (1) 

is  a  sphere. 

The  locus  of  the  equation 

(2>  ?  +  P-?- 

which  contains  only  the  squares  of  the 


tive  coefficient.,  is  called  an  hyperboloid 
of  one  sheet.  The  sections  of  the  sur-. 
face  formed  by  planes  parallel  to  the 
AT-plane  (Rule,  p.  207)  are  ellipses, 
while  those  formed  biplanes  parallel 
to  either  of  the  other  coordinate  planes 
are  conies  of  the  hyperbolic  type.  T\v«  > 
systems  of  straight  lines  lie  .oJL_the 
hyperboloid  and  it  is  therefore  called 
a  ruled  surface. 

The  loci  of  the  equations 


are  also  hyperboloids  of  one  sheet. 


216 


ANALYTIC   GEOMETRY 


The  locus  of  the  equation 


(4) 


y 


with  two  negative  coefficients,  is  called  an  hyperboloid  of  two  sheets.* 

Sections  formed  by  planes  paral- 
lel to  the  rz-plane  (Rule,  p.  207) 
are  conies  of  the  elliptic  type, 
while  those  formed  by  planes 
parallel  to  either  of  the  other 

coordinate  planes  are  hyperbolas. 
The^loci  of  the  equations 


(5) 


X 


,2  ~2                               ~2  ~.2          ^2 

J_   ^L  _  1.  —  1                  ^  l^_  .     ^_  _  1 

.>      I       7  o  o  *J                         .»  70  I          <j    J-. 

a2       &2  c2                    a-  ^»2       c2 


are  also  hyperboloids  of  two  sheets. 
The  locus  of  the  equation 

which  contains  the  squares  of  two  variables  with  positive  coeffi- 
cients and  the  first  power  of  the  third  variable,  is  called  anjelliptic 
paraboloid.  Sections  formed  by  planes  parallel  to  the  AT-plane 
"(Eule,  p.  207)  are  conies  of  the 
elliptic  type,  while  those  formed 
by  planes  parallel  to  either  of  the 
other  coordinate  planes  are  equal 
parabolas. 

The  loci  of  the  equations 


a*       fl- 
are also  elliptic  paraboloids. 


*  The  number  of  sheets  refers  to  the  number  of  separate  parts  of  which  the  surface 
consists. 


PLATE  II 


Elliptic  Paraboloid  Hyperbolic  Paraboloid 

XON-('ENTRAL  QUADRICS 


Hyperboloid  til'  one  sheet 


Hvperbolie  Paraboloid 


SURFACES,  CURVES,  AND  EQUATIONS 


217 


The  locus  of  the  equation 


x 


(8) 


which  differs  from  (6)  in  that  one 
of  the  squares  has  a  'negative 
coefficient,  is  called  an  hyperbolic 
paraboloid.  Sections  formed  by 
planes  parallel  to  the  AT-plane  X' 
(Rule,  p.  207)  are  conies  of  the  hy- 
perbolic type,  while  those  formed 
by  planes  parallel  to  either  of  the 
other  coordinate  planes  are  equal 
parabolas.  There  are  two  systems  of  straight  lines  lying  on  the 
surface. 

The  loci  of  the  equations 


(9) 


are  also  hyperbolic  paraboloids. 

The  ellipsoid  and  the  hyperboloids  are  symmetrical  with 
respect  to  the  origin  (Theorem  IV,  p.  208)  and  are  therefore 
called  central  quadrics,  while  the  paraboloids  are  called  non-central 
quadrics  because  they  have  no  center  of  symmetry. 

PROBLEMS 

1  .  Discuss  and  construct  the  loci  of  the  following  equations. 

(a)  4x2  +  9  7/2  +  16z2  =  144.  (g)  9x2  -  .y*  +  9z*  =  36. 

(b)  4x2  +  9y2  -  16*2  =  144.  (h)  z*  -  4x2  -  4y*  =  Hi. 

(c)  4  x2  -  9  y2  _  16  z*  =  144.  (i)  16  a*  +  y'2  +  16  z*  =  61. 


(d)  x2  +  16  y2  +  z'2  =  64. 

(e)  y2  +  z2  =  4x. 

(f)  y2  -  z2  =  4x. 


(j)  x2  +  y-  -  z*  =  -j:.. 
(k)  9z2  -  4x2  =  288y. 
(1)  16  x2  +  z2  =  64  y. 


2.  Show  how   to  generate  each  of  the  central  quadrics  by  moving  an 
ellipse  whose  axes  are  variable. 


3.  Show  how  to  generate  each  of  the  paraboloids  by  moving  a  parabola. 


rv  /  re 


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T  n  01    inn     o  >*rc                                        General  Library 
SSSSJKS1                                        University  of  California 

(B139s22)4/6                                                         Berkeley 

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UNIVERSITY  OF  CALIFORNIA  LIBRARY 


